copyright © by holt, rinehart and winston. all rights reserved....and ,,, = = )of


PHYSICS IN ACTION

Although the streams of water of the Place

de la Concorde fountain in Paris, France,

seem to flow through transparent pipes,

they are actually fired in highly concentrated

solitary streams. Each of the streams fol-

lows a parabolic path, just like the path a

baseball follows when it is thrown through

the air. In this chapter you will analyze two-

dimensional motion and solve problems

involving objects projected into the air.

• Why does all of the water in a stream followthe same path?

• How does the angle at which the water isfired affect its path?

CONCEPT REVIEW

Displacement (Section 2-1)

Velocity (Section 2-1)

Acceleration (Section 2-2)

Free fall (Section 2-3)

CHAPTER 3

Two-DimensionalMotion andVectors

83Two-Dimensional Motion and VectorsCopyright © by Holt, Rinehart and Winston. All rights reserved.

Copyright © by Holt, Rinehart and Winston. All rights reserved.Chapter 384

SCALARS AND VECTORS

In Chapter 2 our discussion of motion was limited to two directions, forward

and backward. Mathematically, we described these directions of motion with

a positive or negative sign. This chapter explains a method of describing the

motion of objects that do not travel along a straight line.

Vectors indicate direction; scalars do not

Each of the physical quantities we will encounter in this book can be catego-

rized as either a scalar or a vector quantity. A scalar is a quantity that can be

completely specified by its magnitude with appropriate units; that is, a scalar

has magnitude but no direction. Examples of scalar quantities are speed, vol-

ume, and the number of pages in this textbook. A vector is a physical quantity

that has both direction and magnitude.

As we look back to Chapter 2, we can see that displacement is an example

of a vector quantity. An airline pilot planning a trip must know exactly how

far and which way to fly. Velocity is also a vector quantity. If we wish to

describe the velocity of a bird, we must specify both its speed (say, 3.5 m/s)

and the direction in which the bird is flying (say, northeast). Another example

of a vector quantity is acceleration, which was also discussed in Chapter 2.

Vectors are represented by symbols

In physics, quantities are often represented by symbols, such as t for

time. To help you keep track of which symbols represent vector quan-

tities and which are used to indicate scalar quantities, this book will

use boldface type to indicate vector quantities. Scalar quantities are

designated by the use of italics. Thus, to describe the speed of a bird

without giving its direction, you would write v = 3.5 m/s. But a veloc-

ity, which includes a direction, is written as v = 3.5 m/s to the north-

east. Handwritten, a vector can be symbolized by showing an arrow

drawn above the abbreviation for a quantity, such as v�� = 3.5 m/s to

the northeast.

One way to keep track of vectors and their directions is to use

diagrams. In diagrams, vectors are shown as arrows that point in the

direction of the vector. The length of a vector arrow in a diagram is

related to the vector’s magnitude. For example, in Figure 3-1 the

arrows represent the velocities of the two soccer players running

toward the soccer ball.

3-1Introduction to vectors

3-1 SECTION OBJECTIVES

• Distinguish between a scalarand a vector.

• Add and subtract vectorsusing the graphical method.

• Multiply and divide vectorsby scalars.

Figure 3-1The lengths of the vector arrows represent the magnitudes of these two soccer players’ velocities.

scalar

a physical quantity that has onlya magnitude but no direction

vector

a physical quantity that has botha magnitude and a direction


The word vector is also used by air-line pilots and navigators. In thiscontext, a vector is the particularpath followed or to be followed,given as a compass heading.

85Two-Dimensional Motion and Vectors

Because the player on the right of the soccer ball is moving faster, the arrow

representing that velocity is drawn longer than the arrow representing the

velocity of the player on the left.

Vectors can be added graphically

When adding vectors, you must make certain that they have the same units

and describe similar quantities. For example, it would be meaningless to add a

velocity vector to a displacement vector because they describe different physi-

cal quantities. Similarly, it would be meaningless, as well as incorrect, to add

two displacement vectors that are not expressed in the same units (in meters,

in feet, and so on).

Section 2-1 covered vector addition and subtraction in one dimension.

Think back to the example of the gecko that ran up a tree from a 20 cm mark-

er to an 80 cm marker. Then the gecko reversed direction and ran back to the

50 cm marker. Because the two parts of this displacement are opposite, they

can be added together to give a total displacement of 30 cm. The answer found

by adding two vectors in this way is called the resultant.Consider a student walking to school. The student walks 1600 m to a

friend’s house, then 1600 m to the school, as shown in Figure 3-2. The stu-

dent’s total displacement during his walk to school is in a direction from his

house to the school, as shown by the dotted line. This direct path is the vector

sum of the student’s displacement from his house to his friend’s house and his

displacement from the friend’s house to school. How can this resultant dis-

placement be found?

One way to find the magnitude and direction of the

student’s total displacement is to draw the situation to

scale on paper. Use a reasonable scale, such as 50 m on

land equals 1 cm on paper. First draw the vector repre-

senting the student’s displacement from his house to his

friend’s house, giving the proper direction and scaled

magnitude. Then draw the vector representing his walk

to the school, starting with the tail at the head of the first

vector. Again give its scaled magnitude and the right

direction. The magnitude of the resultant vector can

then be determined by using a ruler to measure the

length of the vector pointing from the tail of the first

vector to the head of the second vector. The length of

that vector can then be multiplied by 50 (or whatever

scale you have chosen) to get the actual magnitude of

the student’s total displacement in meters.

The direction of the resultant vector may be deter-

mined by using a protractor to measure the angle

between the first vector and the resultant.

(b)

(a)

(c)

Figure 3-2A student walks from his house to his friend’s house (a), thenfrom his friend’s house to the school (b).The student’s resultantdisplacement (c) can be found using a ruler and a protractor.

resultant

a vector representing the sum oftwo or more vectors


PROPERTIES OF VECTORS

Now consider a case in which two or more vectors act at the same point.

When this occurs, it is possible to find a resultant vector that has the same net

effect as the combination of the individual vectors. Imagine looking down

from the second level of an airport at a toy car moving at 0.80 m/s across a

moving walkway that moves at 1.5 m/s, as graphically represented in Figure3-3. How can you determine what the car’s resultant velocity will look like

from your vantage point?

Vectors can be moved parallel to themselves in a diagram

Note that the car’s resultant velocity while moving from one side of the walk-

way to the other will be the combination of two independent velocities. Thus,

the car traveling for t seconds can be thought of as traveling first at 0.80 m/s

across the walkway for t seconds, then down the walkway at 1.5 m/s for t sec-

onds. In this way, we can draw a given vector anywhere in the diagram as long

as the vector is parallel to its previous alignment and still points in the same

direction. Thus, you can draw one vector with its tail starting at the tip of the

other as long as the size and direction of each vector do not change.

Determining a resultant vector by drawing a vector from the tail of the first

vector to the tip of the last vector is known as the triangle method of addition.

Again, the magnitude of the resultant vector can be measured using a ruler,

and the angle can be measured with a protractor. In the next section, we will

develop a technique for adding vectors that is less time-consuming because it

involves a calculator instead of a ruler and protractor.

Vectors can be added in any order

When two or more vectors are added, the sum is independent of the order of

the addition. This idea is demonstrated by a runner practicing for a marathon

along city streets, as represented in Figure 3-4. In (a) the runner takes one

path during a run, and in (b) the runner takes another. Regardless of which

path the runner takes, the runner will have the same total displacement,

expressed as d. Similarly, the vector sum of two or more vectors is the same

regardless of the order in which the vectors are added, provided that the mag-

nitude and direction of each vector remain the same.

To subtract a vector, add its opposite

Vector subtraction makes use of the definition of the negative of a vector. The

negative of a vector is defined as a vector with the same magnitude as the origi-

nal vector but opposite in direction. For instance, the negative of the velocity

of a car traveling 30 m/s to the west is −30 m/s to the west, or 30 m/s to the

east. Thus, as shown below, adding a vector to its negative vector gives zero.

v + (−v) = 30 m/s + (−30 m/s) = 30 m/s − 30 m/s = 0 m/s

Car

(c)

(a)(b)

vwalkway = 1.5 m/s

vresultant

v car

= 0

.80

m/s

Figure 3-3The resultant velocity (a) of a toycar moving at a velocity of 0.80 m/s(b) across a moving walkway with avelocity of 1.5 m/s (c) can be foundusing a ruler and a protractor.

Figure 3-4A marathon runner’s displacement,d, will be the same regardless of whether the runner takes path(a) or (b).

(a)

d

(b)

d

TOPIC: VectorsGO TO: www.scilinks.orgsciLINKS CODE: HF2031

NSTA

Copyright © by Holt, Rinehart and Winston. All rights reserved.87Two-Dimensional Motion and Vectors

When adding vectors in two dimensions, you can add a negative vector to a

positive vector that does not point along the same line by using the triangle

method of addition.

Multiplying or dividing vectors by scalars results in vectors

There are mathematical operations in which vectors can multiply other vec-

tors, but they are not necessary for the scope of this book. This book does,

however, make use of vectors multiplied by scalars, with a vector as the result.

For example, if a cab driver obeys a customer who tells him to go twice as fast,

that cab’s original velocity vector, vcab, is multiplied by the scalar number 2.

The result, written 2vcab, is a vector with a magnitude twice that of the origi-

nal vector and pointing in the same direction.

On the other hand, if another cab driver is told to go twice as fast in the

opposite direction, this is the same as multiplying by the scalar number −2.

The result is a vector with a magnitude two times the initial velocity but

pointing in the opposite direction, written as −2vcab.

Section Review

1. Which of the following quantities are scalars, and which are vectors?

a. the acceleration of a plane as it takes off

b. the number of passengers on the plane

c. the duration of the flight

d. the displacement of the flight

e. the amount of fuel required for the flight

2. A roller coaster moves 85 m horizontally, then travels 45 m at an angle of

30.0° above the horizontal. What is its displacement from its starting

point? Use graphical techniques.

3. A novice pilot sets a plane’s controls, thinking the plane will fly at

2.50 × 102 km/h to the north. If the wind blows at 75 km/h toward the

southeast, what is the plane’s resultant velocity? Use graphical techniques.

4. While flying over the Grand Canyon, the pilot slows the plane’s engines

down to one-half the velocity in item 3. If the wind’s velocity is still

75 km/h toward the southeast, what will the plane’s new resultant velo-

city be? Use graphical techniques.

5. Physics in Action The water used in many fountains is recycled.

For instance, a single water particle in a fountain travels through an 85 m

system and then returns to the same point. What is the displacement of a

water particle during one cycle?


3-2Vector operations

COORDINATE SYSTEMS IN TWO DIMENSIONS

In Chapter 2, the motion of a gecko climbing a tree was described as motion

along the y-axis. The direction of the displacement of the gecko along the axis

was denoted by a positive or negative sign. The displacement of the gecko can

now be described by an arrow pointing along the y-axis, as shown in Figure 3-5.A more versatile system for diagraming the motion of an object, however,

employs vectors and the use of both the x- and y-axes simultaneously.

The addition of another axis not only helps describe motion in two dimen-

sions but also simplifies analysis of motion in one dimension. For example,

two methods can be used to describe the motion of a jet moving at 300 m/s to

the northeast. In one approach, the coordinate system can be turned so that

the plane is depicted as moving along the y-axis, as in Figure 3-6(a). The jet’s

motion also can be depicted on a two-dimensional coordinate system whose

axes point south to north and west to east, as shown in Figure 3-6(b).The problem with the orientation in the first case is that the axis must be

turned again if the direction of the plane changes. Also, it will become difficult

to describe the direction of another plane that is not traveling exactly north-

east. Thus, axes are often designated with the positive y-axis pointing north

and the positive x-axis pointing east, as shown in Figure 3-6(b).Similarly, when analyzing the motion of objects thrown into the air, orient-

ing the y-axis perpendicular to the ground, and therefore parallel to the direc-

tion of free-fall acceleration, greatly simplifies things.


• Identify appropriate coordi-nate systems for solvingproblems with vectors.

• Apply the Pythagorean theorem and tangent func-tion to calculate the magni-tude and direction of aresultant vector.

• Resolve vectors into compo-nents using the sine andcosine functions.

• Add vectors that are notperpendicular.

∆y

y

Figure 3-5A gecko’s displacement while climb-ing a tree can be represented by anarrow pointing along the y-axis.

v = 300 m/s northeast

yN

S

W E

(a)

Figure 3-6A plane traveling northeast at a velocity of 300 m/s can be repre-sented as either (a) moving along a y-axis chosen to point to thenortheast or (b) moving at an angle of 45° to both the x- and y-axes, which line up with west-east and north-south, respectively.

v = 300 m/s northeast

y

(b)

x


There are no firm rules for applying coordinate systems to situations

involving vectors. As long as you are consistent, the final answer will be correct

regardless of the system you choose. Perhaps the best choice for orienting axes

is the approach that makes solving the problem easiest.

DETERMINING RESULTANT MAGNITUDE AND DIRECTION

In Section 3-1, the magnitude and direction of a resultant were found graphi-

cally by making a drawing. There are, however, drawbacks to this approach: it

is time-consuming, and the accuracy of the answer depends on how carefully

the diagram is drawn and measured. There is a better, simpler method using

the Pythagorean theorem and the tangent function.

Use the Pythagorean theorem to find the magnitude of the resultant

Imagine a tourist climbing a pyramid in Egypt. The tourist knows the height

and width of the pyramid and would like to know the distance covered in a

climb from the bottom to the top of the pyramid.

As can be seen in Figure 3-7, the magnitude of the tourist’s vertical dis-

placement, ∆y, is the height of the pyramid, and the magnitude of the hori-

zontal displacement, ∆x, equals the distance from one edge of the pyramid to

the middle, or half the pyramid’s width. Notice that these two vectors are per-

pendicular and form a right triangle with the displacement, d.As shown in Figure 3-8(a), the Pythagorean theorem states that for any

right triangle, the square of the hypotenuse—the side opposite the right

angle—equals the sum of the squares of the other two sides, or legs.

In Figure 3-8(b), the Pythagorean theorem is applied to find the tourist’s

displacement. The square of the displacement is equal to the sum of the

square of the horizontal displacement and the square of the vertical displace-

ment. In this way, you can find out the magnitude of the displacement, d.

PYTHAGOREAN THEOREM FOR RIGHT TRIANGLES

c2 = a2 + b2

(length of hypotenuse)2 = (length of one leg)2 + (length of other leg)2

d

2∆x

∆y

∆x

Figure 3-7Because the base and height of apyramid are perpendicular, we canfind a tourist’s total displacement,d, if we know the height, ∆y, andwidth, 2∆x, of the pyramid.

∆x

∆ydc

c2 = a2 + b2 d2 = ∆x2 + ∆y2b

a

(a) (b)

Figure 3-8(a) The Pythagorean theoremcan be applied to any right tri-angle. (b) It can also be appliedto find the magnitude of a resul-tant displacement.


Use the tangent function to find the direction of the resultant

In order to completely describe the tourist’s displacement, you must also know

the direction of the tourist’s motion. Because ∆x, ∆y, and d form a right trian-

gle, as shown in Figure 3-9(b), the tangent function can be used to find the

angle q, which denotes the direction of the tourist’s displacement.

For any right triangle, the tangent of an angle is defined as the ratio of the

opposite and adjacent legs with respect to a specified acute angle of a right tri-

angle, as shown in Figure 3-9(a).As shown below, the quantity of the opposite leg divided by the magnitude

of the adjacent leg equals the tangent of the angle.

The inverse of the tangent function, which is shown below, indicates the

angle.

q = tan−1�o

a

p

d

p

j�

DEFINITION OF THE TANGENT FUNCTION FOR RIGHT TRIANGLES

tan q = o

a

p

d

p

j tangent of angle =

o

ad

p

j

p

a

o

c

s

e

i

n

te

t l

l

e

e

g

g

θ

HypotenuseOpposite

Adjacent(a)

(b)

θtan = opp

adj

θtan = ∆y∆x

θ tan−1= ∆y

∆x( )

θ

∆x

∆yd

Figure 3-9(a) The tangent function can beapplied to any right triangle, and (b) it can also be used to find thedirection of a resultant displacement.

SAMPLE PROBLEM 3A

Finding resultant magnitude and direction

P R O B L E MAn archaeologist climbs the Great Pyramid in Giza,Egypt. If the pyramid’s height is 136 m and its width is2.30 � 102 m, what is the magnitude and the directionof the archaeologist’s displacement while climbingfrom the bottom of the pyramid to the top?

S O L U T I O NGiven: ∆y = 136 m ∆x = 1

2(width) = 115 m

Unknown: d = ? q = ?

Diagram: Choose the archaeologist’s starting

position as the origin of the coordinate system.

Choose an equation(s) or situation: The Pythagorean theorem can be used

to find the magnitude of the archaeologist’s displacement. The direction of

the displacement can be found by using the tangent function.

d2 = ∆x2 + ∆y2

tan q = ∆∆

x

y

1. DEFINE

2. PLAN

∆y =136 m

∆x = 115 m

d

x

y

θ


Rearrange the equation(s) to isolate the unknown(s):

d = �∆�x�2�+� ∆�y�2�

q = tan−1�∆∆

x

y�

Substitute the values into the equation(s) and solve:

d = �(1�15� m�)2� +� (�13�6�m�)2�

q = tan−1�113

1

6

5

m

m�

Because d is the hypotenuse, the archaeologist’s displacement should be less

than the sum of the height and half of the width. The angle is expected to be

more than 45° because the height is greater than half of the width.

q = 49.8°

d = 178 m

3. CALCULATE

4. EVALUATE

CALCULATOR SOLUTION

Be sure your calculator is set to cal-culate angles measured in degrees.Most calculators have a buttonlabeled “DRG” that, when pressed,toggles between degrees, radians,and grads.

1. A truck driver attempting to deliver some furniture travels 8 km east,

turns around and travels 3 km west, and then travels 12 km east to his

destination.

a. What distance has the driver traveled?

b. What is the driver’s total displacement?

2. While following the directions on a treasure map, a pirate walks 45.0 m

north, then turns and walks 7.5 m east. What single straight-line dis-

placement could the pirate have taken to reach the treasure?

3. Emily passes a soccer ball 6.0 m directly across the field to Kara, who

then kicks the ball 14.5 m directly down the field to Luisa. What is the

ball’s total displacement as it travels between Emily and Luisa?

4. A hummingbird flies 1.2 m along a straight path at a height of 3.4 m

above the ground. Upon spotting a flower below, the hummingbird

drops directly downward 1.4 m to hover in front of the flower. What is

the hummingbird’s total displacement?

PRACTICE 3A

Finding resultant magnitude and direction


RESOLVING VECTORS INTO COMPONENTS

In the pyramid example, the horizontal and vertical parts that add up to give

the tourist’s actual displacement are called components. The x component is

parallel to the x-axis. The y component is parallel to the y-axis. These compo-

nents can be either positive or negative numbers with units.

Any vector can be completely described by a set of perpendicular compo-

nents. When a vector points along a single axis, as do the quantities in Chapter

2, the second component of the vector is equal to zero.

By breaking a single vector into two components, or resolving it into its

components, an object’s motion can sometimes be described more conve-

niently in terms of directions, such as north to south or east to west.

To illustrate this point, let’s examine a scene on the set of a new action movie.

For this scene a biplane travels at 95 km/h at an angle of 20° relative to the

ground. Attempting to film the plane from below, a camera team travels in a

truck, keeping the truck beneath the plane at all times, as shown in Figure 3-10.How fast must the truck travel to remain directly below the plane?

To find out the velocity that the truck must maintain to stay beneath the

plane, we must know the horizontal component of the plane’s velocity. Once

more, the key to solving the problem is to recognize that a right triangle can

be drawn using the plane’s velocity and its x and y components. The situation

can then be analyzed using trigonometry.

The sine and cosine functions are defined in terms of the lengths of the

sides of such right triangles. The sine of an angle is the ratio of the leg oppo-

site that angle to the hypotenuse.

In Figure 3-11, the leg opposite the 20° angle represents the y component,

vy , which describes the vertical speed of the airplane. The hypotenuse, vplane,is the resultant vector that describes the airplane’s total motion.

The cosine of an angle is the ratio between the leg adjacent to that angle

and the hypotenuse.

In Figure 3-11, the leg adjacent to the 20° angle represents the x component,

vx, which describes the horizontal speed of the airplane. This x component

equals the speed that the truck must maintain to stay beneath the plane.

DEFINITION OF THE COSINE FUNCTION FOR RIGHT TRIANGLES

cos q = h

a

y

d

p

j cosine of an angle =

a

h

d

y

j

p

a

o

c

t

e

e

n

n

t

u

le

se

g

DEFINITION OF THE SINE FUNCTION FOR RIGHT TRIANGLES

sin q = o

h

p

yp

p sine of an angle =

o

h

p

y

p

p

o

o

s

te

it

n

e

u

le

se

g

20˚

vplane

v truck

Figure 3-10A truck carrying a film crew mustbe driven at the correct velocity toenable the crew to film the under-side of a biplane flying at an angle of 20° to the ground at a speed of95 km/h.

vplane = 95 km/h

vx

vy20°

Figure 3-11To stay beneath the biplane, thetruck must be driven with a velocityequal to the x component (vx) ofthe biplane’s velocity.

components of a vector

the projections of a vector alongthe axes of a coordinate system


SAMPLE PROBLEM 3B

Resolving vectors

P R O B L E MFind the component velocities of a helicopter traveling 95 km/h at anangle of 35° to the ground.

S O L U T I O NGiven: v = 95 km/h q = 35°

Unknown: vx = ? vy = ?

Diagram: The most convenient coordinate system is one

with the x-axis directed along the ground and

the y-axis directed vertically.

Choose an equation(s) or situation: Because the axes are

perpendicular, the sine and cosine functions can be used to find

the components.

sin q = v

v

y

cos q = v

v

x


vy = v(sin q)

vx = v(cos q)

Substitute the values into the equation(s) and solve:

vy = (95 km/h)(sin 35°)

vx = (95 km/h)(cos 35°)

Because the component velocities form a right tri-

angle with the helicopter’s actual velocity, the

Pythagorean theorem can be used to check whether

the magnitudes of the components are correct.

v2 = vx2 + vy

2

(95)2 ≈ (78)2 + (54)2

9025 ≈ 9000

The slight difference is due to rounding.

vx = 78 km/h

vy = 54 km/h

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

x

y v = 95 km/h

vx

vy

35°

CALCULATOR SOLUTION

When using your calculator to solve aproblem, perform trigonometric func-tions such as sin, cos, and tan first,before multiplication. This approachwill help you maintain the proper num-ber of significant digits.


ADDING VECTORS THAT ARE NOTPERPENDICULAR

Until this point, the vector-addition problems concerned vectors that are per-

pendicular to one another. However, many objects move in one direction, and

then turn at an acute angle before continuing their motion.

Suppose that a plane initially travels 50 km at an angle of 35° to the

ground, then climbs at only 10° to the ground for 220 km. How can you deter-

mine the magnitude and direction for the vector denoting the total displace-

ment of the plane?

Because the original displacement vectors do not form a right triangle, it is

not possible to directly apply the tangent function or the Pythagorean theo-

rem when adding the original two vectors.

Determining the magnitude and the direction of the resultant can be

achieved by resolving each of the plane’s displacement vectors into their x and

Chapter 394

PRACTICE 3B

Resolving vectors

1. How fast must a truck travel to stay beneath an airplane that is moving

105 km/h at an angle of 25° to the ground?

2. What is the magnitude of the vertical component of the velocity of the

plane in item 1?

3. A truck drives up a hill with a 15° incline. If the truck has a constant

speed of 22 m/s, what are the horizontal and vertical components of the

truck’s velocity?

4. What are the horizontal and vertical components of a cat’s displacement

when it has climbed 5 m directly up a tree?

5. Find the horizontal and vertical components of the 125 m displacement

of a superhero who flies down from the top of a tall building at an angle

of 25° below the horizontal.

6. A child rides a toboggan down a hill that descends at an angle of 30.5° to

the horizontal. If the hill is 23.0 m long, what are the horizontal and ver-

tical components of the child’s displacement?

7. A skier squats low and races down an 18° ski slope. During a 5 s interval,

the skier accelerates at 2.5 m/s2. What are the horizontal (perpendicular

to the direction of free-fall acceleration) and vertical components of the

skier’s acceleration during this time interval?


y components. Then the components along each axis can be added together.

As shown in Figure 3-12, these vector sums will be the two perpendicular

components of the resultant, d. The magnitude of the resultant can then be

found using the Pythagorean theorem, and its direction can be found using

the tangent function.

∆x2

d2

d1 d

∆x1

∆y1

∆y2 Figure 3-12Add the components of the original dis-placement vectors to find two componentsthat form a right triangle with the resul-tant vector.

SAMPLE PROBLEM 3C

Adding vectors algebraically

P R O B L E MA hiker walks 25.5 km from her base camp at 35° south of east. On the sec-ond day, she walks 41.0 km in a direction 65° north of east, at which pointshe discovers a forest ranger’s tower. Determine the magnitude and direc-tion of her resultant displacement between the base camp and theranger’s tower.

S O L U T I O NSelect a coordinate system, draw a sketch of the vectors tobe added, and label each vector.

Figure 3-13 depicts the situation drawn on a coordinate sys-

tem. The positive y-axis points north and the positive x-axis

points east. The origin of the axes is the base camp. In the

chosen coordinate system, the hiker’s direction q1 during

the first day is signified by a negative angle because clock-

wise movement from the positive x-axis is conventionally

considered to be a negative angle.

Given: q1 = −35° q2 = 65° d1 = 25.5 km d2 = 41.0 km

Unknown: d = ? q = ?

Find the x and y components of all vectors.Make a separate sketch of the displacements for each day. The values for

each of the displacement components can be determined by using the sine

and cosine functions. Because the hiker’s angle on the first day is negative,

the y component of her displacement during that day is negative.

sin q = ∆d

y

cos q = ∆d

x

1.

2.

continued onnext page

d1

d2

d

Ranger’s tower

Basecamp

x

y

2θ1θθ

Figure 3-13IN

TERACTIVE

•

T U T O RPHYSICSPHYSICS

Module 2 “Vector Additionand Resolution” provides aninteractive lesson with guidedproblem-solving practice toteach you how to add differentvectors, especially those thatare not at right angles.


d1 = 25.5 km

∆x1

∆y1

x

y

1 = −35°θ

∆y2

∆x2x

y

2 = 65°θ

d 2 =

41.

0 km

Figure 3-15

Figure 3-14

For day 1: ∆x1 = d1 (cos q1) = (25.5 km) [cos (−35°)](Figure 3-14)

∆x1 = 21 km

∆y1 = d1 (sin q1) = (25.5 km) [sin (−35°)]

∆y1 = −15 km

For day 2: ∆x2 = d2 (cos q2) = (41.0 km) (cos 65°)(Figure 3-15)

∆x2 = 17 km

∆y2 = d2 (sin q2) = (41.0 km) (sin 65°)

∆y2 = 37 km

Find the x and y components of the total displacement.First add together the x components to find the total displace-

ment in the x direction. Then perform the same operation for

the y direction.

∆xtot = ∆x1 + ∆x2 = 21 km + 17 km = 38 km

∆ytot = ∆y1 + ∆y2 = −15 km + 37 km = 22 km

Use the Pythagorean theorem to find the magnitude of the resultant vector.Because the components ∆xtot and ∆ytot are perpendicular, the Pythagore-

an theorem can be used to find the magnitude of the resultant vector.

d2 = (∆xtot)2 + (∆ytot)

2

d = �(∆�xt�ot�)2� +� (�∆�yt�ot�)2� = �(3�8�km�)2� +� (�22� k�m�)2�

Use a suitable trigonometric function to find the angle the resultant vec-tor makes with the x-axis.

The direction of the resultant can be found using the tangent function.

q = tan−1�∆∆

x

y�

q = tan−1�∆∆x

yt

t

o

o

t

t� = tan−1�232

8

k

k

m

m�

Evaluate your answer.If the diagram is drawn to scale, compare the algebraic results with the

drawing. The calculated magnitude seems reasonable because the distance

from the base camp to the ranger’s tower is longer than the distance hiked

during the first day and slightly longer than the distance hiked during the

second day. The calculated direction of the resultant seems reasonable

because the angle in Figure 3-13 looks to be about 30°.

q = (3.0 × 101)° north of east

d = 44 km

5.

3.

4.

6.

96 Chapter 3


1. A football player runs directly down the field for 35 m before turning to

the right at an angle of 25° from his original direction and running an

additional 15 m before getting tackled. What is the magnitude and direc-

tion of the runner’s total displacement?

2. A plane travels 2.5 km at an angle of 35° to the ground, then changes

direction and travels 5.2 km at an angle of 22° to the ground. What is the

magnitude and direction of the plane’s total displacement?

3. During a rodeo, a clown runs 8.0 m north, turns 35° east of north, and runs

3.5 m. Then, after waiting for the bull to come near, the clown turns due east

and runs 5.0 m to exit the arena. What is the clown’s total displacement?

4. An airplane flying parallel to the ground undergoes two consecutive dis-

placements. The first is 75 km 30.0° west of north, and the second is

155 km 60.0° east of north. What is the total displacement of the

airplane?

PRACTICE 3C

Adding vectors algebraically

Section Review

1. Identify a convenient coordinate system for analyzing each of the follow-

ing situations:

a. a dog walking along a sidewalk

b. an acrobat walking along a high wire

c. a submarine submerging at an angle of 30° to the horizontal

2. Find the magnitude and direction of the resultant velocity vector for the

following perpendicular velocities:

a. a fish swimming at 3.0 m/s relative to the water across a river that

moves at 5.0 m/s

b. a surfer traveling at 1.0 m/s relative to the water across a wave that is

traveling at 6.0 m/s

3. Find the component vectors along the directions noted in parentheses.

a. a car displaced northeast by 10.0 km (north and east)

b. a duck accelerating away from a hunter at 2.0 m/s2 at an angle of 35°to the ground (horizontal and vertical)

4. Find the resultant displacement of a fox that heads 55° north of west for

10.0 m, then turns and heads west for 5.0 m.


3-3Projectile motion

TWO-DIMENSIONAL MOTION

In the last section, quantities such as displacement and velocity were shown to

be vectors that can be resolved into components. In this section, these compo-

nents will be used to understand and predict the motion of objects thrown

into the air.

Use of components avoids vector multiplication

How can you know the displacement, velocity, and acceleration of a ball at any

point in time during its flight? All of the kinematic equations from Chapter 2

could be rewritten in terms of vector quantities. However, when an object is

propelled into the air in a direction other than straight up or down, the veloc-

ity, acceleration, and displacement of the object do not all point in the same

direction. This makes the vector forms of the equations difficult to solve.

One way to deal with these situations is to avoid using the complicated vec-

tor forms of the equations altogether. Instead, apply the technique of resolving

vectors into components. Then you can apply the simpler one-dimensional

forms of the equations for each component. Finally, you can recombine the

components to determine the resultant.

Components simplify projectile motion

When a long jumper approaches his jump, he runs along a straight line, which

can be called the x-axis. When he jumps, as shown in Figure 3-16, his velocity

has both horizontal and vertical components. Movement in this plane can be

depicted using both the x- and y-axes.

Note that in Figure 3-17(b) the jumper’s velocity vector is resolved into its

two component vectors. This way, the jumper’s motion can be analyzed using

the kinematic equations applied to one direction at a time.


• Recognize examples of pro-jectile motion.

• Describe the path of a pro-jectile as a parabola.

• Resolve vectors into theircomponents and apply thekinematic equations to solveproblems involving projectilemotion.

v

(a)

Figure 3-17(a) A long jumper’s velocity whilesprinting along the runway can berepresented by a single vector.(b) Once the jumper is airborne,the jumper’s velocity at any instantcan be described by the compo-nents of the velocity.

Figure 3-16When the long jumper is in the air,his velocity has both a horizontaland a vertical component.

vx

vy

(b)


In this section, we will focus on the form of two-dimensional motion called

projectile motion. Objects that are thrown or launched into the air and are

subject to gravity are called projectiles. Some examples of projectiles are soft-

balls, footballs, and arrows when they are projected through the air. Even a long

jumper can be considered a projectile.

Projectiles follow parabolic trajectories

The path of a projectile is a curve called a parabola, as shown in Figure 3-18(a). Many people mistakenly believe that projectiles eventually fall straight

down, much like a cartoon character does after running off a cliff. How-

ever, if an object has an initial horizontal velocity in any given time interval,

there will be horizontal motion throughout the flight of the projectile. Note that

for the purposes of samples and exercises in this book, the horizontal velocity of

the projectile will be considered constant. This velocity would not be constant if

we accounted for air resistance. With air resistance, a projectile slows down as it

collides with air particles. Hence, as shown in Figure 3-18(b), the true path of a

projectile traveling through Earth’s atmosphere is not a parabola.

Projectile motion is free fall with an initial horizontal velocity

To understand the motion a projectile undergoes, first examine Figure 3-19.The red ball was dropped at the same instant the yellow ball was launched

horizontally. If air resistance is disregarded, both balls hit the ground at the

same time.

By examining each ball’s position in relation to the horizontal lines and to

one another, we see that the two balls fall at the same rate. This may seem

impossible because one is given an initial velocity and the other begins from

rest. But if the motion is analyzed one component at a time, it makes sense.

First, consider the red ball that falls straight down. It has no motion in the

horizontal direction. In the vertical direction, it starts from rest (vy,i = 0 m/s)

and proceeds in free fall. Thus, the kinematic equations from Chapter 2 can be

applied to analyze the vertical motion of the falling ball. Note that the acceler-

ation, a, can be rewritten as −g because the only vertical component of accel-

eration is free-fall acceleration. Note also that ∆y is negative.

projectile motion

free-fall with an initial horizontalvelocity

Path with air resistance

Path without air resistance

(a)(b)

Figure 3-18(a) Without air resistance, a soccerball being headed into the air wouldbe represented as traveling along aparabola. (b) With air resistance,the soccer ball would travel along ashorter path, which would not be aparabola.

Figure 3-19This is a strobe photograph of twotable-tennis balls released at thesame time. Even though the yellowball is given an initial horizontal veloc-ity and the red ball is simply dropped,both balls fall at the same rate.

TOPIC: Projectile motionGO TO: www.scilinks.orgsciLINKS CODE: HF2032

NSTA


The greatest distance a regulation-size baseball has ever been thrownis 135.9 m, by Glen Gorbous in 1957.

Chapter 3100

Now consider the components of motion of the yellow ball that is launched

in Figure 3-19. This ball undergoes the same horizontal displacement during

each time interval. This means that the ball’s horizontal velocity remains con-

stant (if air resistance is assumed to be negligible). Thus, when using the kine-

matic equations from Chapter 2 to analyze the horizontal motion of a

projectile, the initial horizontal velocity is equal to the horizontal velocity

throughout the projectile’s flight. A projectile’s horizontal motion is described

by the following equation.

Next consider the initial motion of the launched yellow ball in Figure 3-19.Despite having an initial horizontal velocity, the launched ball has no initial

velocity in the vertical direction. Just like the red ball that falls straight down,

the launched yellow ball is in free fall. Its vertical motion is described by the

same free-fall equations. In any time interval, the launched ball undergoes the

same vertical displacement as the ball that falls straight down. This is why

both balls reach the ground at the same time.

To find the velocity of a projectile at any point during its flight, find the

vector sum of the components of the velocity at that point. Use the Pythago-

rean theorem to find the magnitude of the velocity, and use the tangent func-

tion to find the direction of the velocity.

HORIZONTAL MOTION OF A PROJECTILE

vx = vx,i = constant

∆x = vx ∆t

VERTICAL MOTION OF A PROJECTILE THAT FALLS FROM REST

vy,f = −g∆t

vy,f2 = −2g∆y

∆y = − 12

g(∆t)2

Projectile Motion

M A T E R I A L S L I S T

✔ 2 identical balls

✔ slope or ramp

SAFETY CAUTION

Perform this experiment away from wallsand furniture that can be damaged.

Roll a ball off a table. At the instant therolling ball leaves the table, drop a secondball from the same height above the floor.Do the two balls hit the floor at the same

time? Try varying the speed at which youroll the first ball off the table. Does vary-ing the speed affect whether the two ballsstrike the ground at the same time? Nextroll one of the balls down a slope. Dropthe other ball from the base of the slopeat the instant the first ball leaves the slope.Which of the balls hits the ground first inthis situation?



SAMPLE PROBLEM 3D

Projectiles launched horizontally

P R O B L E MThe Royal Gorge Bridge in Colorado rises 321 m above the Arkansas River.Suppose you kick a little rock horizontally off the bridge. The rock hits thewater such that the magnitude of its horizontal displacement is 45.0 m.Find the speed at which the rock was kicked.

S O L U T I O NGiven: ∆y = −321 m ∆x = 45.0 m ay = g = 9.81 m/s2

Unknown: vi = ?

Diagram: The initial velocity vector of the

rock has only a horizontal compo-

nent. Choose the coordinate system

oriented so that the positive y direc-

tion points upward and the positive

x direction points to the right.

Choose the equation(s) or situation:Because air resistance can be neglected, the rock’s

horizontal velocity remains constant.

∆x = vx∆t

Because there is no initial vertical velocity, the following equation applies.

∆y = − 12

g(∆t)2

Rearrange the equation(s) to isolate the unknown(s):Note that the time interval is the same for the vertical and horizontal dis-

placements, so the second equation can be rearranged to solve for ∆t.

∆t =�2

−∆�g

y� where ∆y is negative

vx = ∆∆

x

t = ��

2

−∆�g

y��∆x

Substitute the values into the equation(s) and solve:The value for vx can be either positive or negative because of the square root.

Because the direction was not asked for, use the positive root for v.

vx = �� (45.0 m) =

To check your work, estimate the value of the time interval for ∆x and solve

for ∆y. If vx is about 5.5 m/s and ∆x = 45 m, ∆t ≈ 8 s. If you use an approxi-

mate value of 10 m/s2 for g, ∆y ≈ −320 m, almost identical to the given value.

5.56 m/s−9.81 m/s2

(2)(−321 m)

1. DEFINE

2. PLAN

3. CALCULATE

4. EVALUATE

vx = vxi = vi = ?

ay = g =9.81 m/s2

45.0 m

–321 m

yx


Use components to analyze objects launched at an angle

Let us examine a case in which a projectile is launched at an angle to the hori-

zontal, as shown in Figure 3-20. The projectile has an initial vertical compo-

nent of velocity as well as a horizontal component of velocity.

Suppose the initial velocity vector makes an angle q with the horizontal.

Again, to analyze the motion of such a projectile, the object’s motion must be

resolved into its components. The sine and cosine functions can be used to

find the horizontal and vertical components of the initial velocity.

vx,i = vi (cos q) and vy,i = vi (sin q)

We can substitute these values for vx,i and vy,i into the kinematic equations

from Chapter 2 to obtain a set of equations that can be used to analyze the

motion of a projectile launched at an angle.

As we have seen, the velocity of a projectile launched at an angle to the

ground has both horizontal and vertical components. The vertical motion is

similar to that of an object that is thrown straight up with an initial velocity.

PROJECTILES LAUNCHED AT AN ANGLE

vx = vi (cos q) = constant

∆x = vi(cos q)∆t

vy,f = vi(sin q) − g∆t

vy,f2 = vi

2(sin q)2 − 2g∆y

∆y = vi(sin q)∆t − 12

g(∆t)2

1. An autographed baseball rolls off of a 0.70 m high desk and strikes the

floor 0.25 m away from the base of the desk. How fast was it rolling?

2. A cat chases a mouse across a 1.0 m high table. The mouse steps out of

the way, and the cat slides off the table and strikes the floor 2.2 m from

the edge of the table. What was the cat’s speed when it slid off the table?

3. A pelican flying along a horizontal path drops a fish from a height of

5.4 m. The fish travels 8.0 m horizontally before it hits the water below.

What is the pelican’s initial speed?

4. If the pelican in item 3 was traveling at the same speed but was only

2.7 m above the water, how far would the fish travel horizontally before

hitting the water below?

PRACTICE 3D

Projectiles launched horizontally

vivy,i

vx,i

θ

Figure 3-20An object is projected with an initialvelocity, vi, at an angle of q . Byresolving the initial velocity into itsx and y components, the kinematicequations can be applied todescribe the motion of the projec-tile throughout its flight.

Chapter 3102

PHYSICSPHYSICSModule 3“Two-Dimensional Motion”provides an interactive lessonwith guided problem-solvingpractice to teach you aboutanalyzing motion at angles,including projectile motion.


SAMPLE PROBLEM 3E

Projectiles launched at an angle

P R O B L E MA zookeeper finds an escaped monkey hanging from a light pole. Aimingher tranquilizer gun at the monkey, the zookeeper kneels 10.0 m from thelight pole, which is 5.00 m high. The tip of her gun is 1.00 m above theground. The monkey tries to trick the zookeeper by dropping a banana,then continues to hold onto the light pole. At the moment the monkeyreleases the banana, the zookeeper shoots. If the tranquilizer dart travelsat 50.0 m/s, will the dart hit the monkey, the banana, or neither one?

S O L U T I O NSelect a coordinate system.

As shown in Figure 3-21, the positive

y-axis points up along the tip of the

gun, and the positive x-axis points

along the ground toward the light

pole. Because the dart leaves the gun at

a height of 1.00 m, the vertical distance

to the monkey (and to the banana) is

4.00 m rather than 5.00 m.

Use the tangent function to find the initial angle that the initial velocitymakes with the x-axis.

q = tan−1�∆∆

x

y� = tan−1�41.

0

0

.

0

0

m

m�

q = 21.8°

Choose a kinematic equation to solve for time.First rearrange the equation for motion along the x-axis to isolate the

unknown, ∆t, which is the time it takes the bullet to travel the horizontal

distance from the tip of the gun to the pole.

∆x = vi(cos q)∆t

∆t = vic

∆o

x

sq =

∆t = 0.215 s

Using your knowledge of free fall, find out how far each object will fallduring this time.For the banana:

This is a free-fall problem, where vi = 0.

∆y = − 12

g(∆t)2 = − 12

(9.81 m/s2)(0.215 s)2

∆y = −0.227 m

Thus, the banana will be 0.227 m below its starting point.

10.0 m(50.0 m/s)(cos 21.8°)

1.

2.

3.

4.

continued onnext page

θ

y

x

vi

10.0 m

4.00 m

1.00 m

Figure 3-21


The dart has an initial vertical component of velocity equal to vi(sinq), so:

∆y = vi(sin q)∆t − 12

g(∆t)2

∆y = (50.0 m/s)(sin 21.8°)(0.215 s) − 12

(9.81 m/s2)(0.215 s)2

∆y = 3.99 m − 0.227 m = 3.76 m

The dart will be 3.76 m above its starting point.

Analyze the results.By rearranging our equation for displacement, we can find the final height

of both the banana and the dart.

yf = yi + ∆y

ybanana, f = 5.00 m + (−0.227 m) =

ydart, f = 1.00 m + 3.76 m =

The dart hits the banana. The slight difference is due to rounding.

4.76 m above the ground

4.77 m above the ground

5.

PRACTICE 3E

Projectiles launched at an angle

Chapter 3104

1. In a scene in an action movie, a stuntman jumps from the top of one

building to the top of another building 4.0 m away. After a running start,

he leaps at an angle of 15° with respect to the flat roof while traveling at a

speed of 5.0 m/s. Will he make it to the other roof, which is 2.5 m shorter

than the building he jumps from?

2. A golfer can hit a golf ball a horizontal distance of over 300 m on a good

drive. What maximum height will a 301.5 m drive reach if it is launched

at an angle of 25.0° to the ground? (Hint: At the top of its flight, the ball’s

vertical velocity component will be zero.)

3. A baseball is thrown at an angle of 25° relative to the ground at a speed

of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long

was it in the air? How high was the tallest spot in the ball’s path?

4. Salmon often jump waterfalls to reach their breeding grounds. Starting

2.00 m from a waterfall 0.55 m in height, at what minimum speed must a

salmon jumping at an angle of 32.0° leave the water to continue upstream?

5. A quarterback throws the football to a stationary receiver who is 31.5 m

down the field. If the football is thrown at an initial angle of 40.0° to the

ground, at what initial speed must the quarterback throw the ball for it

to reach the receiver? What is the ball’s highest point during its flight?


Section Review

1. Which of the following are examples of projectile motion?

a. an airplane taking off

b. a tennis ball lobbed over a net

c. a plastic disk sailing across a lawn

d. a hawk diving to catch a mouse

e. a parachutist drifting to Earth

f. a frog jumping from land into the water

2. Which of the following exhibit parabolic motion?

a. a flat rock skipping across the surface of a lake

b. a three-point shot in basketball

c. the space shuttle while orbiting Earth

d. a ball bouncing across a room

e. a cliff diver

f. a life preserver dropped from a stationary helicopter

g. a person skipping

3. An Alaskan rescue plane drops a package

of emergency rations to a stranded party

of explorers, as illustrated in Figure 3-22.The plane is traveling horizontally at

100.0 m/s at a height of 50.0 m above the

ground. What horizontal distance

does the package travel before striking

the ground?

4. Find the velocity (magnitude and direction) of the package in item 3 just

before it hits the ground.

5. During a thunderstorm, a tornado lifts a car to a height of 125 m above

the ground. Increasing in strength, the tornado flings the car horizon-

tally with an initial speed of 90.0 m/s. How long does the car take to

reach the ground? How far horizontally does the car travel before hitting

the ground?

6. Physics in Action Streams of water in a fountain shoot from one

level to the next. A particle of water in a stream takes 0.50 s to travel between

the first and second level. The receptacle on the second level is a horizontal

distance of 1.5 m away from the spout on the first level. If the water is pro-

jected at an angle of 33°, what is the initial speed of the particle?

7. Physics in Action If a water particle in a stream of water in a foun-

tain takes 0.35 s to travel from spout to receptacle when shot at an angle

of 67° and an initial speed of 5.0 m/s, what is the vertical distance

between the levels of the fountain?

50.0 m

vplane = 100.0 m/s

Figure 3-22


3-4Relative motion


• Describe situations in termsof frame of reference.

• Solve problems involving rel-ative velocity.

(a)

(b)

Figure 3-23When viewed from the plane (a), the stunt dummy (representedby the maroon dot) falls straight down. When viewed from a sta-tionary position on the ground (b), the stunt dummy follows aparabolic projectile path.

FRAMES OF REFERENCE

If you are moving at 80 km/h north and a car passes you going 90 km/h, to

you the faster car seems to be moving north at 10 km/h. Someone standing on

the side of the road would measure the velocity of the faster car as 90 km/h

toward the north. This simple example demonstrates that velocity measure-

ments depend on the frame of reference of the observer.

Velocity measurements differ in different frames of reference

Observers using different frames of reference may measure different displace-

ments or velocities for an object in motion. That is, two observers moving with

respect to each other would generally not agree on some features of the motion.

Let us return to the example of the stunt dummy that is dropped from an air-

plane flying horizontally over Earth with a constant velocity. As shown in

Figure 3-23(a), a passenger on the airplane would describe the motion of the

dummy as a straight line toward Earth, whereas an observer on the ground

would view the trajectory of the dummy as that of a projectile, as shown in

Figure 3-23(b). Relative to the ground, the dummy would have a vertical com-

ponent of velocity (resulting from free-fall acceleration and equal to the velocity

measured by the observer in the airplane) and a horizontal component of veloc-

ity given to it by the airplane’s motion. If the airplane continued to move hori-

zontally with the same velocity, the dummy would enter the swimming pool

directly beneath the airplane (assuming negligible air resistance).


Like velocity, displacement andacceleration depend on the frame inwhich they are measured. In somecases, it is instructive to visualizegravity as the ground acceleratingtoward a projectile rather than the projectile accelerating towardthe ground.


1. Elevator acceleration A boy bounces asmall rubber ball in an elevator that is going down. Ifthe boy drops the ball as the elevator is slowingdown, is the ball’s acceleration relative to the eleva-tor less than or greater than its acceleration relativeto the ground?

RELATIVE VELOCITY

The case of the faster car overtaking your car was easy to solve with a mini-

mum of thought and effort, but you will encounter many situations in which

a more systematic method of solving such problems is beneficial. To develop

this method, write down all the information that is given and that you want to

know in the form of velocities with subscripts appended.

vse = +80 km/h north (Here the subscript se means the velocity

of the slower car with respect to Earth.)

vfe = +90 km/h north (The subscript fe means the velocity

of the fast car with respect to Earth.)

We want to know vfs, which is the velocity of the fast car with respect to the

slower car. To find this, we write an equation for vfs in terms of the other

velocities, so on the right side of the equation the subscripts start with f and

eventually end with s. Also, each velocity subscript starts with the letter that

ended the preceding velocity subscript.

vfs = vfe + ves

The boldface notation indicates that velocity is a vector quantity. This

approach to adding and monitoring subscripts is similar to vector addition, in

which vector arrows are placed head to tail to find a resultant.

If we take north to be the positive direction, we know that ves = −vse

because an observer in the slow car perceives Earth as moving south at a

velocity of 80 km/h while a stationary observer on the ground (Earth) views

the car as moving north at a velocity of 80 km/h. Thus, this problem can be

solved as follows:

vfs = +90 km/h − 80 km/h = +10 km/h

The positive sign means that the fast car appears (to the occupants of the

slower car) to be moving north at 10 km/h.

There is no general equation to work relative velocity problems; instead,

you should develop the necessary equations on your own by following the

above technique for writing subscripts.

2. Aircraft carrier Why doesa plane landing on an aircraft carrierapproach the carrier from the rear instead offrom the front?


SAMPLE PROBLEM 3F

Relative velocity

P R O B L E MA boat heading north crosses a wide river with a velocity of 10.00 km/hrelative to the water. The river has a uniform velocity of 5.00 km/h dueeast. Determine the boat’s velocity with respect to an observer on shore.

S O L U T I O NGiven: vbr = 10.00 km/h due north (velocity of the

boat, b, with respect to the river, r)

vre = 5.00 km/h due east (velocity of the

river, r, with respect to Earth, e)

Unknown: vbe = ?

Diagram:

Choose an equation(s) or situation: To find vbe, write the

equation so that the subscripts on the right start with b and end with e.

vbe = vbr + vre

As in Section 3-2, we use the Pythagorean theorem to calculate the magni-

tude of the resultant velocity and the tangent function to find the direction.

(vbe)2 = (vbr)

2 + (vre)2

tan q = v

v

b

re

r


vbe = �(v�br�)2� +� (�vr�e)�2�

q = tan−1�v

v

b

re

r�

Substitute the known values into the equation(s) and solve:

vbe = �(1�0.�00� k�m�/h�)2� +� (�5.�00� k�m�/h�)2�

q = tan−1�150.0.000�

The boat travels at a speed of 11.18 km/h in the direction 26.6° east of north

with respect to Earth.

q = 26.6°

vbe = 11.18 km/h

1. DEFINE

2. PLAN

3. CALCULATE

108 Chapter 3

4. EVALUATE

vre

vbr vbe

x

y

θN

S

W E


PRACTICE 3F

Relative velocity

1. A passenger at the rear of a train traveling at 15 m/s relative to Earth

throws a baseball with a speed of 15 m/s in the direction opposite the

motion of the train. What is the velocity of the baseball relative to Earth

as it leaves the thrower’s hand? Show your work.

2. A spy runs from the front to the back of an aircraft carrier at a velocity of

3.5 m/s. If the aircraft carrier is moving forward at 18.0 m/s, how fast

does the spy appear to be running when viewed by an observer on a

nearby stationary submarine? Show your work.

3. A ferry is crossing a river. If the ferry is headed due north with a speed of

2.5 m/s relative to the water and the river’s velocity is 3.0 m/s to the east,

what will the boat’s velocity be relative to Earth? (Hint: Remember to

include the direction in describing the velocity.)

4. A pet-store supply truck moves at 25.0 m/s north along a highway.

Inside, a dog moves at 1.75 m/s at an angle of 35.0° east of north. What is

the velocity of the dog relative to the road?

Section Review

1. Describe the motion of the following objects if they are observed from

the stated frames of reference:

a. a person standing on a platform viewed from a train traveling north

b. a train traveling north viewed by a person standing on a platform

c. a ball dropped by a boy walking at a speed of 1 m/s viewed by the boy

d. a ball dropped by a boy walking 1 m/s as seen by a nearby viewer who

is stationary

2. A woman on a 10-speed bicycle travels at 9 m/s relative to the ground as

she passes a little boy on a tricycle going in the opposite direction. If the

boy is traveling at 1 m/s relative to the ground, how fast does the boy

appear to be moving relative to the woman? Show your work.

3. A girl at an airport rolls a ball north on a moving walkway that moves

east. If the ball’s speed with respect to the walkway is 0.15 m/s and the

walkway moves at a speed of 1.50 m/s, what is the velocity of the ball

relative to the ground?


In Section 3-4, you learned that velocity measurements are not absolute; every

velocity measurement depends on the frame of reference of the observer with

respect to the moving object. For example, imagine that someone riding a bike

toward you at 25 m/s (v) throws a softball toward you. If the bicyclist measures

the softball’s speed (u�) to be 15 m/s, you would perceive the ball to be moving

toward you at 40 m/s (u) because you have a different frame of reference than

the bicyclist does. This is expressed mathematically by the equation u = v + u�,

which is also known as the classical addition of velocities.

The speed of light

As stated in the “Time dilation” feature in Chapter 2, according to Einstein’s

special theory of relativity, the speed of light is absolute, or independent of all

frames of reference. If, instead of a softball, the bicyclist were to shine a beam

of light toward you, both you and the bicyclist would measure the light’s

speed as 3.0 × 108 m/s. This would remain true even if the bicyclist were

moving toward you at 99 percent of the speed of light. Thus, Einstein’s theory

requires a different approach to the addition of velocities. Einstein’s modifi-

cation of the classical formula, which he derived in his 1905 paper on special

relativity, covers both the case of the softball and the case of the light beam.

u = 1 +

v

(

+

vu

u

�/

�

c2)

In the equation, u is the velocity of an object in a reference frame, u� is the

velocity of the same object in another reference frame, v is the velocity of one

reference frame relative to another, and c is the speed of light.

The universality of Einstein’s equation

How does Einstein’s equation cover both cases? First we shall consider the

bicyclist throwing a softball. Because c2 is such a large number, the vu�/c2

term in the denominator is very small for velocities typical of our everyday

experience. As a result, the denominator of the equation is nearly equal to 1.

Hence, for speeds that are small compared with c, the two theories give nearly

the same result, u = v + u�, and the classical addition of velocities can be used.

111

Table 3-1 Classical and relativistic addition of velocities

c � 2.997 925 84 � 108 m/s Classical Relativisticaddition addition

Speed between Speed in Speed in Speed inframes (v) A (u�) B (u) B (u)

000 025 m/s 000 0 15 m/s 000 000 040 m/s 000 000 040 m/s

100 000 m/s 100 000 m/s 000 200 000 m/s 000 200 000 m/s

50% of c 50% of c 299 792 584 m/s 239 834 067 m/s

90% of c 90% of c 539 626 651 m/s 298 136 271 m/s

99.99% of c 99.99% of c 599 525 210 m/s 299 792 582 m/s

Figure 3-24According to Einstein’s relativisticequation for the addition of velo-cities, material particles can neverreach the speed of light.

However, when speeds approach the speed of light, vu�/c2 increases, and

the denominator becomes greater than 1 and less than 2. When this occurs,

the difference between the two theories becomes significant. For example, if a

bicyclist moving toward you at 80 percent of the speed of light were to throw

a ball to you at 70 percent of the speed of light, you would observe the ball

moving toward you at about 96 percent of the speed of light rather than the

150 percent of the speed of light predicted by classical theory. In this case, the

difference between the velocities predicted by each theory cannot be ignored,

and the relativistic addition of velocities must be used.

In this last example, it is significant that classical addition predicts a speed

greater than the speed of light (1.5c), while the relativistic addition predicts a

speed less than the speed of light (0.96c). In fact, no matter how close the

speeds involved are to the speed of light, the relativistic equation yields a

result less than the speed of light, as seen in Table 3-1.How does Einstein’s equation cover the second case, in which the bicyclist

shines a beam of light toward you? Einstein’s equation predicts that any

object traveling at the speed of light (u� = c) will appear to travel at the speed

of light (u = c) for an observer in any reference frame:

u = 1 +

v

(

+

vu

u

�/

�

c2) =

1 +

v

(

+vc

c

/c2) =

1 +v +

(v

c

/c) =

(c

v

++v

c

)/c = c

This corresponds with our earlier statement that the bicyclist measures the

beam of light traveling at the same speed that you do, 3.0 × 108 m/s, even

though you have a different reference frame than the bicyclist does. This

occurs regardless of how fast the bicycle is moving because v (the bicycle’s

speed) cancels from the equation. Thus, Einstein’s relativistic equation suc-

cessfully covers both cases. So, Einstein’s equation is a more general case of

the classical equation, which is simply the limiting case.

TOPIC: Speed of lightGO TO: www.scilinks.orgsciLINKS CODE: HF2033

NSTA

Two-Dimensional Motion and VectorsCopyright © by Holt, Rinehart and Winston. All rights reserved.

Chapter 3112

KEY IDEAS

Section 3-1 Introduction to vectors• A scalar is a quantity completely specified by only a number with appropri-

ate units, whereas a vector is a quantity that has magnitude and direction.

• Vectors can be added graphically using the triangle method of addition, in

which the tail of one vector is placed at the head of the other. The resultant is

the vector drawn from the tail of the first vector to the head of the last vector.

Section 3-2 Vector operations• The Pythagorean theorem and the tangent function can be used to find

the magnitude and direction of a resultant vector.

• Any vector can be resolved into its component vectors using the sine and

cosine functions.

Section 3-3 Projectile motion• Neglecting air resistance, a projectile has a constant horizontal velocity

and a constant downward free-fall acceleration.

• In the absence of air resistance, projectiles follow a parabolic path.

Section 3-4 Relative motion• If the frame of reference is denoted with subscripts (vab is the velocity of a

with respect to b), then the velocity of an object with respect to a different

frame of reference can be found by adding the known velocities so that the

subscript starts with the letter that ends the preceding velocity subscript,

vab = vac + vcb.• If the order of the subscripts is reversed, there is a change in sign, for exam-

ple, vcd = −vdc.

CHAPTER 3Summary

KEY TERMS

components of a vector (p. 92)

projectile motion (p. 99)

resultant (p. 85)

scalar (p. 84)

vector (p. 84)

Variable symbols

Quantities Units

d (vector) displacement m meters

v (vector) velocity m/s meters/second

a (vector) acceleration m/s2 meters/second2

∆x (scalar) horizontal component m meters

∆y (scalar) vertical component m meters

Diagram symbols

displacement vector

velocity vector

acceleration vector

resultant vector

component



VECTORS AND THE GRAPHICALMETHOD

Review questions

1. The magnitude of a vector is a scalar. Explain thisstatement.

2. If two vectors have unequal magnitudes, can theirsum be zero? Explain.

3. What is the relationship between instantaneousspeed and instantaneous velocity?

4. What is another way of saying −30 m/s west?

5. Is it possible to add a vector quantity to a scalarquantity? Explain.

6. Vector A is 3.00 units in length and points along thepositive x-axis. Vector B is 4.00 units in length andpoints along the negative y-axis. Use graphicalmethods to find the magnitude and direction of thefollowing vectors:

a. A + Bb. A − Bc. A + 2Bd. B − A

7. Each of the displacement vectors A and B shown inFigure 3-25 has a magnitude of 3.00 m. Graphicallyfind the following:

a. A + Bb. A − Bc. B − Ad. A − 2B

Figure 3-25x

y

3.00 m

3.00 m

30.0°

A

B

8. A dog searching for a bone walks 3.50 m south, then8.20 m at an angle of 30.0° north of east, and finally15.0 m west. Use graphical techniques to find thedog’s resultant displacement vector.

9. A man lost in a maze makes three consecutive dis-placements so that at the end of the walk he is backwhere he started, as shown in Figure 3-26. The firstdisplacement is 8.00 m westward, and the second is13.0 m northward. Use the graphical method tofind the third displacement.

Conceptual questions

10. If B is added to A, under what conditions does theresultant have the magnitude equal to A + B?

11. Give an example of a moving object that has a veloc-ity vector and an acceleration vector in the samedirection and an example of one that has velocityand acceleration vectors in opposite directions.

12. A student accurately uses the method for combin-ing vectors. The two vectors she combines havemagnitudes of 55 and 25 units. The answer that shegets is either 85, 20, or 55. Pick the correct answer,and explain why it is the only one of the three thatcan be correct.

13. If a set of vectors laid head to tail forms a closedpolygon, the resultant is zero. Is this statement true?Explain your reasoning.

Figure 3-26

X

CHAPTER 3Review and Assess



25. A shopper pushing a cart through a store moves40.0 m south down one aisle, then makes a 90.0°turn and moves 15.0 m. He then makes another90.0° turn and moves 20.0 m. Find the shopper’stotal displacement. Note that you are not given thedirection moved after any of the 90.0° turns, sothere could be more than one answer.(See Sample Problem 3A.)

26. A submarine dives 110.0 m at an angle of 10.0° belowthe horizontal. What are the horizontal and verticalcomponents of the submarine’s displacement? (See Sample Problem 3B.)

27. A person walks 25.0° north of east for 3.10 km.How far would another person walk due north anddue east to arrive at the same location?(See Sample Problem 3B.)

28. A roller coaster travels 41.1 m at an angle of 40.0°above the horizontal. How far does it move hori-zontally and vertically?(See Sample Problem 3B.)

29. A person walks the path shown in Figure 3-27. Thetotal trip consists of four straight-line paths. At theend of the walk, what is the person’s resultant dis-placement measured from the starting point?(See Sample Problem 3C.)

Figure 3-27

PROJECTILE MOTION

Review questions

30. A bullet is fired horizontally from a pistol, andanother bullet is dropped simultaneously from thesame height. If air resistance is neglected, whichbullet hits the ground first?

31. If a rock is dropped from the top of a sailboat’smast, will it hit the deck at the same point whetherthe boat is at rest or in motion at constant velocity?

N

S

W E300.0 m

100.0 m

200.0 m

150.0 m30.0°60.0°

?

VECTOR OPERATIONS

Review questions

14. Can a vector have a component equal to zero andstill have a nonzero magnitude?

15. Can a vector have a component greater than itsmagnitude?

16. Explain the difference between vector addition andvector resolution.

17. If the component of one vector along the directionof another is zero, what can you conclude aboutthese two vectors?

18. How would you add two vectors that are not per-pendicular or parallel?

Conceptual questions

19. If A + B = 0, what can you say about the compo-nents of the two vectors?

20. Under what circumstances would a vector havecomponents that are equal in magnitude?

21. The vector sum of three vectors gives a resultantequal to zero. What can you say about the vectors?

Practice problems

22. A girl delivering newspapers travels three blockswest, four blocks north, then six blocks east.

a. What is her resultant displacement? b. What is the total distance she travels?

(See Sample Problem 3A.)

23. A golfer takes two putts to sink his ball in the holeonce he is on the green. The first putt displaces the ball6.00 m east, and the second putt displaces it 5.40 msouth. What displacement would put the ball in thehole in one putt? (See Sample Problem 3A.)

24. A quarterback takes the ball from the line of scrim-mage, runs backward for 10.0 yards, then runs side-ways parallel to the line of scrimmage for 15.0 yards.At this point, he throws a 50.0-yard forward passstraight down the field. What is the magnitude ofthe football’s resultant displacement?(See Sample Problem 3A.)


32. Does a ball dropped out of the window of a movingcar take longer to reach the ground than onedropped at the same height from a car at rest?

33. A rock is dropped at the same instant that a ball atthe same elevation is thrown horizontally. Whichwill have the greater velocity when it reachesground level?

Practice problems

34. The fastest recorded pitch in Major League Baseballwas thrown by Nolan Ryan in 1974. If this pitchwere thrown horizontally, the ball would fall 0.809 m (2.65 ft) by the time it reached home plate,18.3 m (60 ft) away. How fast was Ryan’s pitch?(See Sample Problem 3D.)

35. A shell is fired from the ground with an initial speedof 1.70 × 103 m/s (approximately five times thespeed of sound) at an initial angle of 55.0° to thehorizontal. Neglecting air resistance, find

a. the shell’s horizontal rangeb. the amount of time the shell is in motion

(See Sample Problem 3E.)

36. A person standing at theedge of a seaside cliffkicks a stone over theedge with a speed of 18m/s. The cliff is 52 mabove the water’s surface,as shown in Figure 3-28.How long does it take forthe stone to fall to thewater? With what speeddoes it strike the water?(See Sample Problem 3D.)

37. A spy in a speed boat is being chased down a riverby government officials in a faster craft. Just as theofficials’ boat pulls up next to the spy’s boat, bothboats reach the edge of a 5.0 m waterfall. If the spy’sspeed is 15 m/s and the officials’ speed is 26 m/s,how far apart will the two vessels be when they landbelow the waterfall?

38. A place kicker must kick a football from a point36.0 m (about 40.0 yd) from the goal. As a result ofthe kick, the ball must clear the crossbar, which is3.05 m high. When kicked, the ball leaves theground with a speed of 20.0 m/s at an angle of 53°to the horizontal.

a. By how much does the ball clear or fall shortof clearing the crossbar?

b. Does the ball approach the crossbar while stillrising or while falling?

(See Sample Problem 3E.)

39. A daredevil is shot out of a cannon at 45.0° to thehorizontal with an initial speed of 25.0 m/s. A net ispositioned at a horizontal distance of 50.0 m fromthe cannon from which it is shot. At what heightabove the cannon should the net be placed in orderto catch the daredevil? (See Sample Problem 3E.)

40. When a water gun is fired while being held horizon-tally at a height of 1.00 m above ground level, thewater travels a horizontal distance of 5.00 m. Achild, who is holding the same gun in a horizontalposition, is also sliding down a 45.0° incline at aconstant speed of 2.00 m/s. If the child fires the gunwhen it is 1.00 m above the ground and the watertakes 0.329 s to reach the ground, how far will thewater travel horizontally?(See Sample Problem 3E.)

41. A ship maneuvers to within 2.50 × 103 m of anisland’s 1.80 × 103 m high mountain peak and fires aprojectile at an enemy ship 6.10 × 102 m on the otherside of the peak, as illustrated in Figure 3-29. If theship shoots the projectile with an initial velocity of2.50 × 102 m/s at an angle of 75.0°, how close to theenemy ship does the projectile land? How close (ver-tically) does the projectile come to the peak?(See Sample Problem 3E.)

Figure 3-29

x

y

vi = +18 m/s

h = 52 mg

Figure 3-28

vi =1.80 × 103 m

75.0°

2.50 × 103 m

2.50 × 102 m/s

6.10 × 102 m


50. A hunter wishes to cross a river that is 1.5 km wideand that flows with a speed of 5.0 km/h. The hunteruses a small powerboat that moves at a maximumspeed of 12 km/h with respect to the water. What isthe minimum time necessary for crossing?(See Sample Problem 3F.)

51. A swimmer can swim in still water at a speed of9.50 m/s. He intends to swim directly across a riverthat has a downstream current of 3.75 m/s.

a. What must the swimmer’s direction be?b. What is his velocity relative to the bank?

(See Sample Problem 3F.)

MIXED REVIEW PROBLEMS

52. A motorboat heads due east at 12.0 m/s across a riverthat flows toward the south at a speed of 3.5 m/s.

a. What is the resultant velocity relative to anobserver on the shore?

b. If the river is 1360 m wide, how long does ittake the boat to cross?

53. A ball player hits a home run, and the baseball justclears a wall 21.0 m high located 130.0 m fromhome plate. The ball is hit at an angle of 35.0° to thehorizontal, and air resistance is negligible. Assumethe ball is hit at a height of 1.0 m above the ground.

a. What is the initial speed of the ball?b. How much time does it take for the ball to

reach the wall?c. Find the velocity components and the speed

of the ball when it reaches the wall.

54. A daredevil jumps a canyon 12 m wide. To do so, hedrives a car up a 15° incline.

a. What minimum speed must he achieve toclear the canyon?

b. If the daredevil jumps at this minimum speed,what will his speed be when he reaches theother side?

55. A 2.00 m tall basketball player attempts a goal 10.00 m from the basket (3.05 m high). If he shootsthe ball at a 45.0° angle, at what initial speed musthe throw the basketball so that it goes through thehoop without striking the backboard?

RELATIVE MOTION

Review questions

42. Explain the statement, “All motion is relative.”

43. What is a frame of reference?

44. When we describe motion, what is a common frameof reference?

45. A small airplane is flying at 50 m/s toward the east.A wind of 20 m/s toward the east suddenly begins toblow, giving the plane a velocity of 70 m/s east.

a. Which of these are component vectors?b. Which is the resultant? c. What is the magnitude of the wind velocity?

46. A ball is thrown upward in the air by a passenger ona train that is moving with constant velocity.

a. Describe the path of the ball as seen by thepassenger. Describe the path as seen by a sta-tionary observer outside the train.

b. How would these observations change if thetrain were accelerating along the track?

Practice problems

47. The pilot of a plane measures an air velocity of165 km/h south. An observer on the ground sees theplane pass overhead at a velocity of 145 km/h towardthe north. What is the velocity of the wind that isaffecting the plane?(See Sample Problem 3F.)

48. A river flows due east at 1.50 m/s. A boat crosses theriver from the south shore to the north shore bymaintaining a constant velocity of 10.0 m/s duenorth relative to the water.

a. What is the velocity of the boat as viewed byan observer on shore?

b. If the river is 325 m wide, how far downstreamis the boat when it reaches the north shore?


49. The pilot of an aircraft wishes to fly due west in a50.0 km/h wind blowing toward the south. The speedof the aircraft in the absence of a wind is 205 km/h.

a. In what direction should the aircraft head?b. What should its speed be relative to the ground?



56. A ball is thrown straight upward and returns to thethrower’s hand after 3.00 s in the air. A second ball isthrown at an angle of 30.0° with the horizontal. Atwhat speed must the second ball be thrown so that itreaches the same height as the one thrown vertically?

57. An escalator is 20.0 m long. If a person stands onthe escalator, it takes 50.0 s to ride from the bottomto the top.

a. If a person walks up the moving escalatorwith a speed of 0.500 m/s relative to the esca-lator, how long does it take the person to getto the top?

b. If a person walks down the “up” escalator withthe same relative speed as in item (a), howlong does it take to reach the bottom?

58. A ball is projected horizontally from the edge of atable that is 1.00 m high, and it strikes the floor at apoint 1.20 m from the base of the table.

a. What is the initial speed of the ball? b. How high is the ball above the floor when its

velocity vector makes a 45.0° angle with thehorizontal?

59. How long does it take an automobile traveling 60.0 km/h to become even with a car that is travel-ing in another lane at 40.0 km/h if the cars’ frontbumpers are initially 125 m apart?

60. The eye of a hurricane passes over Grand BahamaIsland. It is moving in a direction 60.0° north of westwith a speed of 41.0 km/h. Exactly three hours later,the course of the hurricane shifts due north, and itsspeed slows to 25.0 km/h, as shown in Figure 3-30.How far from Grand Bahama is the hurricane 4.50 hafter it passes over the island?

Figure 3-30

41.0 km/h

25.0 km/h

60.0°

N

S

EW

61. A car is parked on a cliff overlooking the ocean onan incline that makes an angle of 24.0° below thehorizontal. The negligent driver leaves the car inneutral, and the emergency brakes are defective. Thecar rolls from rest down the incline with a constantacceleration of 4.00 m/s2 and travels 50.0 m to theedge of the cliff. The cliff is 30.0 m above the ocean.

a. What is the car’s position relative to the baseof the cliff when the car lands in the ocean?

b. How long is the car in the air?

62. A boat moves through a river at 7.5 m/s relative tothe water, regardless of the boat’s direction. If thewater in the river is flowing at 1.5 m/s, how longdoes it take the boat to make a round trip consistingof a 250 m displacement downstream followed by a250 m displacement upstream?

63. A golf ball with an initial angle of 34° lands exactly240 m down the range on a level course.

a. Neglecting air friction, what initial speedwould achieve this result?

b. Using the speed determined in item (a), findthe maximum height reached by the ball.

64. A water spider maintains an average position on thesurface of a stream by darting upstream (against thecurrent), then drifting downstream (with the cur-rent) to its original position. The current in thestream is 0.500 m/s relative to the shore, and thewater spider darts upstream 0.560 m (relative to aspot on shore) in 0.800 s during the first part of itsmotion. Use upstream as the positive direction.

a. Determine both the velocity of the water spiderrelative to the water during its dash upstreamand its velocity during its drift downstream.

b. How far upstream relative to the water doesthe water spider move during one cycle of thisupstream and downstream motion?

c. What is the average velocity of the water spiderrelative to the water for one complete cycle?

65. A car travels due east with a speed of 50.0 km/h.Rain is falling vertically with respect to Earth. Thetraces of the rain on the side windows of the carmake an angle of 60.0° with the vertical. Find thevelocity of the rain with respect to the following:

a. the carb. Earth


68. A science student riding on a flatcar of a train mov-ing at a constant speed of 10.0 m/s throws a balltoward the caboose along a path that the studentjudges as making an initial angle of 60.0° with thehorizontal. The teacher, who is standing on theground nearby, observes the ball rising vertically.How high does the ball rise?

69. A football is thrown toward a receiver with an initialspeed of 18.0 m/s at an angle of 35.0° above the hor-izontal. At that instant, the receiver is 18.0 m fromthe quarterback. In what direction and with whatconstant speed should the receiver run to catch thefootball at the level at which it was thrown?

66. A shopper in a department store can walk up a sta-tionary (stalled) escalator in 30.0 s. If the normallyfunctioning escalator can carry the standing shop-per to the next floor in 20.0 s, how long would ittake the shopper to walk up the moving escalator?Assume the same walking effort for the shopperwhether the escalator is stalled or moving.

67. If a person can jump a horizontal distance of 3.0 mon Earth, how far could the person jump on themoon, where the free-fall acceleration is g/6 andg = 9.81 m/s2? How far could the person jump onMars, where the acceleration due to gravity is 0.38g?

Graphing calculatorsRefer to Appendix B for instructions on download-

ing programs for your calculator. The program

“Chap3” allows you to analyze a graph of height

versus time for a baseball thrown straight up.

Recall the following equation from your studies

of projectiles launched at an angle.

∆y = vi(sin q)∆t − 12

g(∆t)2

The program “Chap3” stored on your graphing

calculator makes use of the projectile motion equa-

tion. Given the initial velocity, your graphing calcu-

lator will use the following equation to graph the

height (Y1) of the baseball versus the time interval

(X) that the ball remains in the air. Note that the

relationships in this equation are the same as those

in the projectile motion equation shown above.

Y1 = VX − 4.9X2

a. The two equations above differ in that the lat-

ter does not include the factor sin q . Why has

this factor been disregarded in the second

equation?

Execute “Chap3” on the p menu, and press

e to begin the program. Enter the value for the

initial velocity (shown below), and press e to

begin graphing.

The calculator will provide the graph of the

displacement function versus time. The x value

corresponds to the time interval in seconds, and the

y value corresponds to the height in meters. If the

graph is not visible, press w and change the set-

tings for the graph window.

Press ◊, and use the arrow keys to trace

along the curve to the highest point of the graph.

The y value there is the greatest height that the ball

reaches. Trace the curve to the right, where the

y value is zero. The x value is the duration of the

ball’s flight.

For each of the following initial velocities,

identify the maximum height and flight time of a

baseball thrown vertically.

b. 25 m/s

c. 35 m/s

d. 75 m/s

e. Does the appearance of the graph represent

the actual trajectory of the ball? Explain.

Press @ q to stop graphing. Press e to

input a new value or ı to end the program.


70. A rocket is launched at an angle of 53° above thehorizontal with an initial speed of 75 m/s, as shownin Figure 3-31. It moves for 25 s along its initial lineof motion with an overall acceleration of 25 m/s2.At this time its engines fail and the rocket proceedsto move as a free body.

a. What is the rocket’s maximum altitude?b. What is the rocket’s total time of flight?c. What is the rocket’s horizontal range? Figure 3-31

a = 25 m/s2

vi = 75 m/s

53°

Performance assessment1. Work in cooperative groups to analyze a game of

chess in terms of displacement vectors. Make a

model chessboard, and draw arrows showing all the

possible moves for each piece as vectors made of

horizontal and vertical components. Then have two

members of your group play the game while the

others keep track of each piece’s moves. Be prepared

to demonstrate how vector addition can be used to

explain where a piece would be after several moves.

2. Use a garden hose to investigate the laws of projectile

motion. Design experiments to investigate how the

angle of the hose affects the range of the water stream.

(Assume that the initial speed of water is constant and

is determined by the pressure indicated by the faucet’s

setting.) What quantities will you measure, and how

will you measure them? What variables do you need

to control? What is the shape of the water stream?

How can you reach the maximum range? How can

you reach the highest point? Present your results to

the rest of the class and discuss the conclusions.

3. How would a physics expert respond to the follow-

ing suggestions made by three airline executives?

Write a script of the expert’s response for perfor-

mance in front of the class.

Airline Executive A: Since the Earth rotates from

west to east, we could operate “static flights”—heli-

copters that begin by hovering above New York City

could begin their landing four hours later, when San

Francisco arrives below.

Airline Executive B: This could work for one-way

flights, but the return trip would take 20 hours.

Airline Executive C: That will never work. It’s like

when you throw a ball up in the air; it comes back to

the same point.

Airline Executive A: That’s only because the Earth’s

motion is not significant during that short a time.

Portfolio projects4. You are helping NASA engineers design a basketball

court for a colony on the moon. How do you antici-

pate the ball’s motion compared with its motion on

Earth? What changes will there be for the players—

how they move and how they throw the ball? What

changes would you recommend for the size of the

court, the basket height, and other regulations in

order to adapt the sport to the moon’s low gravity?

Create a presentation or a report presenting your

suggestions, and include the physics concepts

behind your recommendations.

5. There is conflicting testimony in a court case. A

police officer claims that his radar monitor indicated

that a car was traveling at 176 km/h (110 mi/h). The

driver argues that the radar must have recorded the

relative velocity because he was only going 88 km/h

(55 mi/h). Is it possible that both are telling the

truth? Could one be lying? Prepare scripts for expert

witnesses, for both the prosecution and the defense,

that use physics to justify their positions before the

jury. Create visual aids to be used as evidence to sup-

port the different arguments.

Alternative Assessment

Chapter 3120

VELOCITY OF A PROJECTILE

In this experiment, you will determine the horizontal velocity of a projectile

and compare the effects of different inclined planes on this projectile motion.

PREPARATION

1. Read the entire lab, and plan what measurements you will take.

2. Prepare a data table in your lab notebook with five columns and nine

rows. In the first row, label the columns Trial, Height of ramp (m), Length

of ramp (m), Displacement ∆x (m), Displacement ∆y (m). In the first col-

umn, label the second through ninth rows 1, 2, 3, 4, 5, 6, 7, and 8.

3. Set up the inclined plane at any angle, as shown in Figure 3-32(a). Tape the

aluminum to the end of the plane and to the table. Leave at least 5 cm

between the bottom end of the inclined plane and the edge of the table.

4. In front of the table, place the box to catch the ball after it bounces. Perform

a practice trial to find the correct placement of the box. Use masking tape to

secure the box to the floor. Cover the floor with white paper. Cover the white

paper with the carbon paper, carbon side down. Tape the paper down.

5. Use a piece of cord and tape to hang a washer from the edge of the table so

that the washer hangs a few centimeters above the floor. Mark the floor

directly beneath the washer with tape. Move the washer to another point

on the table edge, and repeat. Connect the two marks on the floor with

masking tape.

PROCEDURE

6. Use tape to mark a starting line near the top of the inclined plane. Mea-

sure the height of the ramp from the tabletop to the tape mark. Measure

the length along the ramp from the tape mark to the tabletop. Measure

CHAPTER 3Laboratory Exercise

OBJECTIVES

•Measure the velocity ofprojectiles in terms ofthe horizontal displace-ment during free fall.

•Compare the velocityand acceleration of pro-jectiles accelerateddown different inclinedplanes.

MATERIALS LIST✔ aluminum sheet, edges

covered with heavy tape✔ C-clamp✔ cardboard box✔ cord✔ inclined plane✔ several sheets carbon paper✔ several large sheets white

paper✔ masking tape✔ meterstick✔ packing tape✔ small metal ball✔ small metal washer✔ support stand and clamp✔ towel or cloth

SAFETY

• Tie back long hair, secure loose clothing, and remove loose jewelry toprevent their getting caught in moving or rotating parts.

• Perform this experiment in a clear area. Falling or dropped massescan cause serious injury.



the distance from the top of the tabletop to the floor.

Enter these values in the data table for Trial 1 as Height

of ramp (m), Length of ramp (m), and Displacement ∆y.

7. Place the metal ball on the inclined plane at the tape

mark. Keep the area around the table clear of people and

obstructions. Release the ball from rest so that it rolls

down the inclined plane, off the table onto the carbon

paper, and bounces into the box.

8. Lift the carbon paper. There should be a heavy carbon

mark on the white paper where the ball landed. Label

the mark with the trial number.

9. Replace the carbon paper and repeat this procedure as Trial 2.

10. With the inclined plane in the same position, place another tape

mark about halfway down the inclined plane. Measure and record

the height and length of the inclined plane from this mark. Use this

mark as the starting point for Trial 3 and Trial 4. Record all data.

11. Raise or lower the inclined plane and repeat the procedure. Perform

two trials for each tape mark as Trials 5, 6, 7, and 8. For each trial,

measure the distance from the carbon mark to the tape line. Record

this distance as Displacement ∆x.

12. Clean up your work area. Put equipment away safely so it is ready to

be used again. Recycle or dispose of used lab materials as directed by your

teacher.

ANALYSIS AND INTERPRETATION

Calculations and data analysis

1. Organizing data Find the time interval for the ball’s motion from the

edge of the table to the floor using the equation for the vertical motion of

a projectile from page 100, where ∆y is the Displacement ∆y recorded in

your data table. The result is the time interval for each trial.

2. Organizing data Using the time interval from item 1 and the value for

Displacement ∆x, calculate the average horizontal velocity for each trial

during the ball’s motion from the edge of the table to the floor.

Conclusions

3. Inferring conclusions What is the relationship between the height of

the inclined plane and the horizontal velocity of the ball? Explain.

4. Inferring conclusions What is the relationship between the length of

the inclined plane and the horizontal velocity of the ball? Explain.

Figure 3-32Step 3: Use tape to cover the sharpedges of the aluminum sheet beforetaping it to the end of the plane. Thealuminum keeps the ball from bounc-ing as it rolls onto the table.Step 4: Place a soft cloth in the boxas shown. Throughout the lab, becareful not to have too much contactwith the carbon paper.Step 5: This tape line will serve as the baseline for measuring thehorizontal distance traveled by theprojectile.

(a)

(b)


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