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Name: _____________________

Period: _____________________

Copyright © 2010, C. Stephen Murray cstephenmurray.com

Trigonometry Basics

Calculators and Trig You will have to find the sin, cos, or tan ratios OR you will have to find θ,

given the ratios. You must know how to use your calculator to do this.

Given the angle (θ), find sin, cos, or tan. Ex: sin 30° = ______

Scientific Calculator: Type “30”

Push “sin”

Answer: “0.5”

: sin 30° = 0.5Answer

Finding Unknowns The real power of trigonometry is that it relates the angles and sides of a right triangle.

If, for example, you know θ and a side, you can find all the other parts of the triangle.

Step 3: Solve

25º

60 cm Y

X

Problem: Find the

length of X.

Variables:

θ = 25º

opp. = Y

adj. = X

hyp.= 60 cm

Step 1:

Assign Variables

Step 2:

Choose a Formula

You know hyp and

need adj., so use cos.

adj.cosθ =

hyp.

adj.cosθ =

hyp.

Xcos25° =

60cm

X.9063 =

60cm

(60).9063 = X

X = 54.4cm

Sin (Sine) − ratio of the opposite side

to the hypotenuse.

Cos (Cosine) − ratio of the adjacent

side to the hypotenuse.

Tan (Tangent) − ratio of the opposite

side to the adjacent side.

Trigonometric Functions

oppSin θ =

hyp

adjCos θ =

hyp

oppTan θ =

adj

30º

60º 16 m

8 m

13.9 m

opp. 13.9mSin60° = = = 0.866

hyp. 16m

opp. 8mSin30° = = = 0.5

hyp. 16m

opp. 8mTan30° = = = .5774

adj. 13.9m

Trigonometric Ratios

for This Triangle

Sin, cos, and tan are ratios, telling you how big

(what percentage) one side is in relation to another.

Scientific Calculator: Press the DRG key

until degrees shows

in the display.

Recheck!

DRG MODE

Graphing Calculator: Push “sin”

Type “30”

Answer: “0.5”

Graphing Calculator: Press the MODE key. Find

where RADIANS is selected.

Select DEGREES instead.

Press ENTER. Recheck!

: θ = 25Answer °

Given sin, cos, or tan, find θ.

Scientific Calculator: Type “.4226”

Push “INV” then “sin”

Gives “25”

Graphing Calculator: Push “2nd” then “sin”

Type “.4226”

Gives “25”

-1

Ex: sinθ = .4226

θ = sin (.4226)

θ = ____

Use Degrees not Radians! When using sin, cos,

and tan your calculator MUST be in degrees or all

of your numbers will be wrong.

Quick Check: The sin 30º = 0.5!

If sin 30º ≠ 0.5, your calculator in in radians

and must be changed.

Caution!

Basic Terms

θ (theta) − variable for any angle.

Hypotenuse − longest side of a triangle.

Opposite − side opposite the angle (θ).

Adjacent − side next to the angle (θ).

θ1

Y

X

θ2 R

Y is opposite to θ1;

Y is adjacent to θ2.

X is adjacent to θ1; X is opposite to θ2.

R is the hypotenuse

for both θ1 and θ2. Which side is opposite?

It depends on the angle.

a

b

c

Pythagorean Theorem

Remember that

a2 + b2 = c2

Where a and b are

either of the sides and

c is the hypotenuse.

Name: _____________________

Period: _____________________


Using your calculator, give the following ratios.

Given the following ratios, use your calculator to find θ.

Cos θ = .8192;

θ = _________

Sin θ = 0.5

θ = _________

Tan θ = .8391

θ = _________

Sin θ = .866

θ = __________

Tan θ = 1.732

θ = __________

Cos θ = 0.5

θ = __________

Tan θ = 1

θ = __________

Cos θ = .866

θ = __________

Sin θ = .7071

θ = __________

Cos 30º = ___________

Tan 15º = ___________

Sin 60º = ___________

Tan 30º = ___________

Sin 45º = ___________

Tan 85º = __________

Cos 45º = __________

Cos 60º = __________

Sin 30º = ___________

Tan 45º = __________

40º

15 cm 9.6 cm

11.5 cm

θ = _________________

Opposite = ___________

Adjacent = ___________

Hypotenuse = ________

75º 8 in 2

in

7.7 in

θ = _________________

Opposite = __________

Adjacent = __________

Hypotenuse = ________

Opposite for 30º = ________

Hypotenuse for 60º = ______

Adjacent for 60º = ________


30º

60º 20 m 10 m

17.3 cm

If opposite = 17.3 cm, then θ = __________ If adjacent = 10 m, then θ = _____________ If adjacent = 17.3 cm, then θ = __________

70º

20º

120 mm 10 mm

17.3 mm







If opposite = 17.3 cm, then θ = ________

30º

10 cm Y

Obviously Y ≠ 9.88 cm.

What went wrong?

Ysin30° =

10cm

Y.988 =

10cm

Y = 9.88cm

35º

Y 26 cm

Step 1: Assign Variables

26 cm = _____________

35º = ________________

Y = _________________

Step 2: Choose a Formula (Sin, Cos, or Tan?)

Step 3: Solve

Y = __________

45º

6 ft

X

Following the three steps at the left,

find the length of X.

How long is Y?

.8660 35º

What is the sin of 75º?

If tan θ = 0.868, solve for θ.

12 cm

? 13 cm

Two sides of this triangle are

given. Calculate the third side.

Trigonometry Basics-2

Name: _____________________

Period: _____________________


Name: _____________________

Period: _____________________


Vector Basics

We use arrows to represent

vectors because vectors

have both magnitude (size)

and direction (which way

it points).

When adding vectors graphi-

cally, put the arrows head to

tail. The resultant points

from start to finish. In this

example your total displace-

ment is the straight line

distance between your initial

and final position NOT the

distance you traveled.

The components are the portions of the vector in the x or y direction, like coordinates on a graph.

Components retain

the units of their vector (and

vice-versa).

vector =

6 m/s

30o

x – component

= 6m/s(cos 30o)

= 5.2 m/s

y – component

= 6m/s(sin 30o)

= 3 m/s

If the vector was a plane, think of the x-component as a race

car trying to stay beneath the plane on the ground. The

y-component could be how fast the plane gains altitude.

start

final

Resultant

Total

displacement

Vector 1

Displacement 1

Vector 2

Displacement 2

The result of adding together two

or more vectors is called a resultant.

Order doesn’t matter when adding

vectors. The resultant will be the same.

V1 V2

Different order: same resultant.

start

finish

V1

V2 R1+2

V1

V2 R1+2

start

finish

o

o

ysin 30 =

40 m

y = (40m)sin 30

(40m)0.5 =y = 20 m

The components tell

you that you would

have to move 34.6 m

in the x-direction and

20 m in the y-direction

to move 40 m at 30º

o

o

xcos 30 =

40 m

x = (40m)cos 30

x = (40m )(.866) = 34.6 m

x – component = 34.6 m

40 m

30o

V1 + V1 = 2V1

V1 –V1

Subtracting vectors: add its opposite

(the negative of the vector).

Multiplying vectors: multiply the size of the vector.

Opposite of V1

Twice the size of V1

Adding Graphically

Components

y –

component

= 20 m

Components can be negative or zero.

50o

75 m

-x component

+y c

om

po

nent

130o 90o

+y c

om

po

nent

No x comp.

100% of

this vector

is vertical

Think directions NOT

angles. Your calculator

will give you positives and

negatives automatically IF

you give the calculator

correct directions.

Units Math and Vectors

Use 130° and your calculator

will give you the correct

+ and − components.

An example of no x-component: a cat climbs up a tree.

The cat doesn’t move horizontally, just vertically.

These angles are

equal (20º), but

point in different

directions.

0º

270º or -90º

90º

180º 20º

20º

180 + 20

= 200º

For your

calculator

only this is 20º.

20º

For your

calculator this

is -20º.

20º

90 + 20

= 110º

X-comp = Hyp.(cosθ) Y-comp = Hyp.(sinθ)

X and Y Components:

Magnitude (length) of the vector

Direction of the vector

Setting up vectors.

Before you calculate

components always

find direction of θ,

as shown above.

30º

90º

180º

-90º

0º

15m The angle is 30º, but the

direction is 150º (180−30).

This vector is 15 m at 150º

7 m/s

7 m/s

45° Same magnitude;

different directions.

4 m

6 m

Same direction;

different magnitudes.

-30º

-30º

Mass is a scalar,

which only requires

magnitude. You

don’t need the direc-

tion of the mass. 2 kg

Name: _____________________

Period: _____________________


1. Resolve

2. Magnitude

3. Resultant

4. Component

5. Direction

6. Vector

A. The portion of the vector on the

x or y axis.

B. To find the x– or y-component of

a vector.

C. The size of a vector (“35” of “35 m”).

D. Tells where a vector is pointing or the

angle of the vector.

E. What you find by adding two vectors

together.

F. Something that has magnitude and

direction.

V1

V2

R A.

V2

V1

R C.

V2

V1

R D.

V2

V1

R B.

7. In figures A—D, which

vectors are added correctly?

If wrong, why?

A.

B.

C.

D.

A. 42 m/s

35o

B.

20 m/s2

45o C.

70o

72 m

30o

20 m/s

V1 V3 V2 V4 V5 Using the vectors at the right, draw the resultants for

the following operations.

8. V2 –V5 =

9. 2V2 + V4 =

10. V3 + 2V4 − V5 =

11. 2V1 − 2V4 =

12. Add 2V1 +V4 mathematically.

13. If each of the vectors is 10º from

the closest axis, determine the

directions of each of the vectors.

θA =

θB =

θC =

θD =

θE =

θF =

θG =

θH =

15. A person walks 12 m across a room.

A. What is their horizontal component?

B. What is their vertical component?

16. Resolve this vector into its components.

17. Resolve these vectors into their components.

Vector Basics -2

0º

270º or -90º

90º

180º

B C

A

G

H E

F

D

14. Find calculator directions for the following vectors.

25º 45º

A. θ = B. θ = C. θ =

Name: _____________________

Period: _____________________


Name: _____________________

Period: _____________________


start

V1 =

V2 =

finish

1. Find the following information for vector 1.

A. How far does vector 1 move horizontally?

(This is the X-component.) X1 =

B. What is the Y-component of vector 1? Y2 =

C. How long is vector 1? (Find the magnitude of

vector 1.)

2. Resolve vector 2 into its x and y components.

(Do the same as in #1)

A. X2 =

B. Y2 =

C. Magnitude of vector 2 =

Each grid square represents 1 m. In this example, you may count squares.

3. Draw the resultant from the start to the finish

(Label it “R”).

4. Add together X1 and X2 =

(this is Xtotal)

5. Ytotal =

6. Using Xtotal and Ytotal, calculate the length of R

(with Xtotal and Ytotal you have two sides of a right triangle).

Ex 1

1. Find the following information for vector 1.

A. How far does vector 1 move horizontally?

(This is the X-component.) X1 =

B. What is the Y-component of vector 1? Y2 =

C. How long is vector 1? (Find the magnitude of

vector 1.)

2. Resolve vector 2 into its x and y components.

(Do the same as in #1)

A. X2 =

B. Y2 =


Each grid square represents 1 m. In this example, you may count squares.

3. Draw the resultant from the start to the finish

(Label it “R”).

4. Add together X1 and X2 =

(this is Xtotal)

5. Ytotal =

6. Using Xtotal and Ytotal, calculate the length of R

(with Xtotal and Ytotal you have two sides of a right triangle).

start

V1 =

V2 =

finish Ex 1

Name: _____________________

Period: _____________________


start V1 =

V2 =

finish

1. A. X1 = B. Y2 =

C. Magnitude of V1 =

2. A. X2 = B. Y2 =


3. Draw the resultant from the start to the finish.

4. A. Xtotal = B. Ytotal =

5. Calculate the magnitude of R.

6. Using X total and Y total, calculate the direction of R.

(see “Adding Vector” notes.)

Ex 2

start V1 =

V2 =

finish

1. A. X1 = B. Y1 =

C. Magnitude of V1 =

2. A. X2 = B. Y2 =


3. Draw the resultant from the start to the finish.

4. A. Xtotal = B. Ytotal =

5. Calculate the magnitude of R.

6. Using X total and Y total, calculate the direction of R.

(see “Adding Vector” notes.)

Ex 2

Name: _____________________

Period: _____________________


Name: _____________________

Period: _____________________


Vx1 = 30.3 m

Vx2 = 25 m

Example: A person walks 35 m at 30° then 50 m

at 60°. Calculate the person’s total displacement.

Step 1. Resolve vectors into their components.

Step 3. Draw a resultant triangle

with xtotal and ytotal . Then, calculate the

resultant’s magnitude and direction.

Use the Pythagorean

Theorem to find the

resultant’s magnitude.

2 2 2

total total

2 2 2

R = x + y

R =55.3 + 60.8 = 6754.73

R = 6754.73 = 82.2 m

Use inverse tangent to find

the resultant’s direction.

total

total

-1

yopp.tan θ =

adj. x

60.8tan θ =

55.3

60.8θ = tan = 47.7

55.3

=

�

Step 2. Calculate xtotal and ytotal by adding up all x-components

and all y-components.

Be sure to keep

track of negatives!

xtotal = 55.3 m

yto

tal =

60

.8 m

= 8

2.2

m

θ = 47.7o

R can be found graphically

(by drawing and measuring)

or mathematically.

v 1 = 35 m

30o

Vy1 =

17.5 m

v 2 =

50

m

60o

Vy2 =

43.3 o

2

2

Vy = 50(sin60 )

Vy = 43.3 m

o

2

2

Vx = 50(cos60 )

Vx = 25 m

o

1

1

Vy = 35(sin30 )

Vy = 17.5 mo

1

1

Vx = 35(cos30 )

Vx = 30.3 m

o

o

ysin120 =

45 m

45(sin120 ) = y

45(.866) = y

y = 39 m

o

o

xcos120 =

45 m

45(cos120 ) = x

All angles

MUST start

at the

+ x axis

D =

45 m

120o

y =

39 m

x = -22.5 m

60o

From

+ x axis

+x

Setting Up Individual Vectors

In this example all components are positive. By taking

your angle from the +x-axis, sine and cosine will give

you positive and negative components automatically.

Answer: R = 82.2 m at 47.7º

Before you begin, be sure all of your

vectors have the same units and all

angles start at the +x-axis. Then, your

calculator will automatically calculate

any positive or negative components.

Direction =

Res

ulta

nt (t

otal

displ

acem

ent)

Res

ulta

nt (t

otal

displ

acem

ent)

= 30.3 + 25 = 55.3 m

x1

x2

xtotal = x1 + x2

yto

tal =

y1 +

y2

...

= 4

3.3

+ 1

7.5

= 6

0.8

m

y1

y2

2

2

tota

l

tota

l

x

+ y

Mag

nitu

de (R

) =

-1 total

total

yθ = tan

x

Where H is the vector,

θ is the angle from the + x axis,

x is the x component of H,

and y is the y-component of H.

These 2 equations

work for any vector

(even if the vector

is vertical

[θ = 90º or 270º]

or horizontal

[θ = 0º or 180º]) .

x = Hcosθ and y = Hsinθ

X and Y Components

45(-0.5) = x

x = -22.5 m

Using 120°

gave a –x

component!

Tan Can’t See –X Tan can’t see the difference

between a negative x and

a negative y. Tan only

gives angles between +90°

and –90.°, so it will never

give you an angle in

quadrants 2 or 3.

( )

( )

1 1

1 1

-4tan tan 2 63.4

2

Tan sees these as the same!

4tan tan 2 63.4

-2

− −

− −

= − = − °

= − = − °

If Xtotal is negative add 180 degrees

1 10.4tan 60

-6

− = °

Since Xtotal is negative:

θ = 60 + 180 = 120º

But we know that this

direction is greater than 90º.

Yto

tal =

10

.4 m

Xtotal = -6 m

θ

90º

0º Which we can

see is true.

Adding Vectors

Name: _____________________

Period: _____________________


3. Resolve the following vectors into their components.

A. C.

25º

80 m 32º

40º

12 m/s

B.

6.5 m

1. A drag racer moves

350 m down a race track.

2. A person walks

150 m north.

4. Find the resultant of the following two vectors.

5. Add these vectors together. Assume each square = 1 m.

6. A person walks 30 m north, then 50 m at 35º. Find their total displacement.

A. Draw the resultant from

the start of the first vector

to the end of the second.

Label it “R”.

B. Xtotal =

C. Ytotal =

40º

X1 = 9.2 m

Y1 =

7.7 m

X2 = −5.2 m

V2 = 6 m

30º

Y2 =

3 m

V 1 =

12 m

start

finish

D. Calculate the magnitude

(length) of R.

E. Calculate the direction

of R.

V2

V1

A. X1 =

Y1 =

B. X2 =

Y2 =

C. Xtotal =

Ytotal =

D. Draw the resultant (R).

E. Calculate R’s magnitude.

F. Calculate R’s direction.

X =

Y =

X =

Y =

A. Below draw R from the start of

V1 to the end of V2. B. Resolve v1 and v2 into their

components (Step 1 on the front)

X1 =

Y2 =

Xtotal = X2 =

Y1 = Ytotal =

Magnitude = Direction =

35º V2

= 50 m

V1 =

30

m

7. Add these vectors: V1 = 55 m at 38o and V2 = 10 m at 26 0o.

A. Draw V1, V2, and R below.

B. Resolve v1 and v2 into their

components (Step 1 on the front)

X1 =

Y2 =

X2 =

Y1 =

Xtotal =

Ytotal =

Magnitude =

Direction =

Adding Vectors -2

Name: _____________________

Period: _____________________


Name: _____________________

Period: _____________________


Ex. 1—Add these vectors: 35 m at 30° and 60 m at 160°.

Additional Adding Vector Examples

D1 = 35 m

30o

x1 = 30.3 m

y1 =

17.5 m

160o

x2 = -56.4 m

y2 =

20.5 m

D2 = 60 m

x1

y1

x2

y2

= 30.3 − 56.4 = −26.1 m

yto

tal =

y1 +

y2

...

= 2

0.5

+ 1

7.5

= 3

8 m

R

xtotal = x1 + x2

2 2 2

total total

2 2 2

R = x + y

R =26.1 + 38

R =46.1 m

total

total

-1

yopp.tan θ =

adj. x

38tan θ =

26.1

38θ = tan = 56.6

-26.1

56.6 180 124.4

=

−

−

− + =

�

R =

46.1 m

θ = 124.4°

xtotal = −26.1m

yto

tal =

38

m 2

2

total

total

x

+ y

-1 total

total

yθ = tan

x

Add 180° to

your angle be-

cause it is obvi-

ously not in the

4th quadrant.


Step 3. Calculate R’s magnitude and direction.

Step 2. Calculate xtotal and ytotal

12

50

m/s

at

90

°

2600 m/s at 320°

R

y1 =

1250

m/s

x2 = 1992 m/s

y2 =

−1670

m/s

R

y1

x2

y2

ytotal = y1 + y2 ... = 1250 −1670

= −421 m

xtotal = x1 + x2

= 1992

total

total

-1

yopp.tan θ =

adj. x

421tan θ =

1992

421θ = tan = 12

1992

=

−

− −

�

xt = 1992m

R

2 2 2

total total

2 2 2

R = x + y

R = 1992 + 421

R =2036 m

yt = −421 m

Ex. 2—Add these vectors: 1250 m/s at 90° and 2600 m at 320°.

These additional examples are for

students to visualize the process of

adding vectors. It is assumed that

students can already calculate

components.


Step 3. Calculate R’s magnitude and direction.

Step 2. Calculate xtotal and ytotal

θ2 = 320º

Name: _____________________

Period: _____________________


Remember that the resultant is drawn from the start of the first vector to the end of the last vector. Not all motion is possible

Relative motion describes an object’s

motion relative (in relation) to different

frames of reference.

In the picture at the right the tortoise has a

speed of 0.1 m/s relative to the truck, but has

a speed of 5.1 m/s relative to the road.

In describing relative motion you must

always give your frame of reference

(the truck or the road).

Vtortoise to truck = 0.1 m/s

Vtruck to road = 5 m/s

Vtortoise to road = 5.1 m/s

X and Y are Independent

Imagine a person walking 1m/s across a

2 m wide railroad car. If the car is at rest

the person will take 2 seconds to cross.

start

end Vperson

= 1 m/s

Vtrain = 0 m/s

2 m

If the railcar is moving, the person still takes 2 seconds to cross the car.

The velocity of the railcar is irrelevant because the velocity of the car

is in the x-direction and the person is moving in the y-direction.

The x and y directions are independent of each other!

start

end Vtrain

Vperson

Vtotal

To calculate Vtotal, use Pythagorean theorem:

Vtotal2 = Vperson

2 + Vtrain2

In the example below, a boat encounters a strong current

in a river. If the boat captain aims straight across the

river, the river will push the boat downstream.

landing River

7 m/s

Boat

5 m/s

VB

VR

VT

launch

landing

Boat 5 m/s

River 7 m/s

launch

But what if the captain wishes to land at a point

straight across the river? The boat will have to

be aimed up stream, but at what angle?

The total velocity of the boat is found by adding the

vectors, which here requires only Pythagorean theorem,

because these vectors are already horizontal and vertical .

A vector triangle shows that this is not possible

because the hypotenuse is smaller than one of the

sides. The boat is not able to go straight across.

VT

VB =

5 m/s

VR = 7 m/s

Desired

path

Relative motion can be solved graphically or mathematically, just like all other vectors. Vectors Still Add

VWind = 20 m/s

45º

V Plane =

40

m/s

A plane flying has a velocity of 40 m/s, relative

to the ground, experiences a west wind that

is moving 20 m/s, relative to the ground.

What is the velocity of

the plane relative to

the ground?

VWind = 20 m/s

V Plane =

40

m/s

V total

Mathematical Solution:

Vxwind = 20

Vxplane = 40(cos45º) = 28.3

Vxtotal = 20 + 28.3 = 48.3m/s

Vywind = 0 (all horizontal)

Vyplane = 40(sin45º) = 28.3

Vytotal = 0 + 28.3 = 28.3m/s

2 248.3 + 28.3

56 m/s

28.3θ = tan-1

48.3

θ = 30.4

total

total

V

V

=

=

°Graphically:

See notes: “Adding Vectors”

Relative Motion

Name: _____________________

Period: _____________________


1. An moving walkway at the airport has a velocity of 2 m/s to the right.

A person walks at a steady pace of 3 m/s.

A. If the person is walking to the right, what is their

velocity relative to the walkway?

B. What is their velocity relative to the ground?

C. How long would it take them to travel to the food

court, 100 m away?

D. How long would it take them to walk back if

they have to walk on the same walkway?

3 m/s

2 m/s

E. How long would it take them to walk to the food

court and back without using the walkway?

2. A toy plane’s is flying 55° going 8 m/s. If the wind is pushing with a velocity of 3 m/s at 30º,

find the total velocity and direction of plane.

3. A boat is traveling 6 m/s at an angle of −30o . The water has a current

flowing 3 m/s directly south . Find the boat’s total velocity and direction.

4. A person can swim 4 m/s. The river has a current flowing 6 m/s directly east.

A. What will be the direction and velocity of the person if they aim directly across the river (north)?

B. If the person swims at constant speed, how long does it take them to swim across the 40 m wide river?

C. How far downstream will the person drift?

D. At what direction will the person have to swim to reach a point directly across the river?

E. If the river’s current increases (is faster), will the person take more or less time to cross the river?

Relative Motion -2

Name: _____________________

Period: _____________________


Hypo. can’t be smaller than a side.

Even if they swam directly

upstream they would be

pushed downstream. They

have to be swimming faster

than the river to get straight

across. Then the hypo is big-

ger than both sides.

The x and y directions are independent of each other. The x-direction speed of

the river has no effect on a person swimming in the y-direction.

Name: _____________________

Period: _____________________


Vi

Vxi = Vicosθ

Vyi =

Visinθ Vx

Vy

In the x-direction the object is at constant speed. So ax = 0 m/s2 and S = D/T.

In the y-direction it is in

freefall. So ay = –9.8 m/s2.

θ

At the top: Vy = 0 m/s,

so V = Vx

Ground

Vyf =

–Vyi stops

The object stops in x-direction

when it hits the ground in

the y-direction, so: ty = tx

Vxi = Vicosθ

A projectile is any thrown, shot, or launched object: a rock; a bullet; a volleyball; a person jumping. All

projectiles follow a parabolic path. The distance a projectile travels in the x-direction is known as its range.

Just as with any vector, the x and y-components of

the initial velocity can be found using sine and cosine.

You know that if you throw a rock from up on a cliff, it

will go farther than if you were on the ground. This is

because the higher up you are, the longer the rock is in

the air. Flight time is dependent on the y-direction only!

3 m

6 m

An object launched

from higher up stays in the

air longer and goes farther.

Time Comes From the Y-direction

Projectile Motion

If you took several

quick pictures of a

projectile in flight you

would notice that in the

y-direction it looks just

like freefall—being

pulled back to the

earth by gravity.

In the x-direction

you would see that the

object is at constant

speed, moving the same

distance each second.

To find the projectile’s

velocity and angle at any

point: calculate Vx and Vy,

then use Pythagorean

Theorem and inverse tan.

Vyi =

V(sinθ)

Vxi = V(cos θ)

θ

Vi

(Initial horizontal velocity)

(Initial vertical

velocity)

Initial velocity

(velocity it was

thrown or shot)

Vyi =

V(sinθ)

If an object is launched

horizontally, the initial

vertical velocity is 0 m/s.

Vi = Vxi

Vyi = 0 m/s

y−direction

ay = −9.8 m/s2

Vi = 0 m/s

Vf = not used

∆y = −8 m

t = ?

Solving Projectile Motion Problems

Use your “Freefall”

notes to help you assign

the y-direction variables.

Just as in all freefall

problems, you will use

one of the kinematic

equations to calculate

unknowns.

Often you will

solve for time in

the y-direction.

The accelerations will

ALWAYS be –9.8 m/s2

and 0 m/s2.

Any projectile motion problem can be easily solved if you : 1) draw the situation;

2) write out what you know in the x and y-directions; 3) solve for unknowns.

9 m/s

8 m

∆x (range) = ?

2

2

2

2

1( ) ( )

2

18 (0 ) ( 9.8)( )

2

8 4.9

81.63

4.9

1.63 1.28 sec

iy v t a t

t t

t

t

t

∆ = +

− = + −

− = −

−= =

−

= =

DS =

T

D = ST

D = 9(1.28)

D = 11.5 m

x−direction

ax = 0 m/s2

Vi = 9 m/s

Vf = 9 m/s

∆x =

t = 1.28 sec

Because it’s

at constant

velocity

From the

y-direction

Since the x-

direction is at

constant speed,

you can ALWAYS

use S = D/T for

your x-direction

equation.

Since horizontal:

Vy = 0 m/s

Name: _____________________

Period: _____________________


Pro

ject

ile

Mo

tio

n E

xa

mp

le

Pro

ble

m:

A c

an

no

n f

ires

a c

an

no

nb

all

. I

f

the

can

no

n’s

an

gle

is

35

o to

th

e g

rou

nd

an

d

the

mu

zzle

vel

oci

ty i

s 5

0 m

/s,

wh

at

is t

he

ran

ge

of

the

can

no

nb

all

?

Ste

p 1

: R

eso

lve

V i

nto

Vx

an

d V

y

θ Vi

VX

i

VY

i V

Yi =

V(s

in θ

)=

28

.7m

/s

VX

i =

V(c

os θ)

= 4

1 m

/s

Vi = 5

0 m

/s

θ =

35

o

X-c

om

po

nen

t

VX

= V

(co

s θ)

VX

= 5

0m

/s(c

os

35

o)

VX

= 4

1m

/s

Y-c

om

po

nen

t

VY

= V

(sin

θ)

VY

= 5

0m

/s(s

in 3

5o)

VY

= 2

8.7

m/s

Vi = 5

0 m

/s

θ =

35

o

∆X

= ?

Ste

p 2

: F

ind

tim

e fr

om

th

e y-

dir

ecti

on

.

ay =

–9

.8 m

/s2

Vi

= V

y =

Vsi

nθ

= 2

8.7

m/s

(se

e st

ep 1

)

Vf

= -

Vi

= -

28

.7m

/s

∆Y

= 0

m (

fro

m g

rou

nd

to

gro

un

d)

∆t

= ?

Ch

oo

se o

ne

of

the

kin

em

ati

c eq

ua

tio

ns:

In t

he

y-d

irec

tio

n,

the

canno

nb

all

is i

n

free

fall

.

Wri

te t

he

va

ria

ble

s y

ou

kn

ow

:

() 2

1(

)2

1 2

if

fi

i

yv

vt

vv

at

yv

ta

t

∆=

+

=+

∆

=+

Ste

p 3

: F

ind

∆x

fro

m t

he

x-d

irec

tio

n.

In t

he

x-d

irec

tio

n,

the

canno

nb

all

trav

els

at c

onst

ant

vel

oci

ty:

Vx,

the

x-c

om

po

nent

of

the

vel

oci

ty.

The

tim

e o

f fl

igh

t w

e ju

st

fou

nd

in t

he

y-d

irec

tio

n.

Wri

te t

he

va

ria

ble

s y

ou

kn

ow

:

ax =

0 m

/s2

Vi =

Vf

Vx =

Vco

sθ =

41

m/s

(se

e st

ep 1

)

t =

5.8

sec

(fr

om

y [

see

step

2])

∆x =

?

∆x =

23

7.8

m

By

kn

ow

ing

on

ly t

he

lau

nch

vel

oci

ty a

nd

an

gle

, yo

u c

an

ca

lcu

late

d t

hat

the

can

no

n-

ba

ll t

rave

led

237

.8 m

eter

s in

5.8

sec

on

ds.

land

ing

Muz

zle

velo

city

()

2

22

1 2 2

f

fi

yv

ta

t

vv

ay

∆=

−

=+

∆

DS

T= 4

1(5

.8)

DS

T

DS

T

D

= =

=

Sin

ce a

x =

0 m

/s2

RA

NG

E

This

is

the

mo

st b

asic

of

all

pro

ject

ile

mo

tio

n p

rob

lem

s: f

rom

the

gro

und

-to

-the-

gro

und

(A

to

E).

S

tep

s 1

and

3 w

ill

AL

WA

YS

be

the

sam

e.

Ste

p 2

is

just

fre

efall

and

wil

l ch

an

ge

acco

rdin

g t

o t

he

situ

atio

n.

Sin

ce y

ou

ha

ve

all

of

the

va

ria

ble

s, y

ou

ca

n

use

an

y f

orm

ula

. S

o u

se t

he

easi

est

on

e:

vf =

vi +

(a

t)

Put

in w

hat

yo

u k

no

w:

-28

.7 =

(28

.7)

+ (

-9.8

t)

-57

.4

= -

9.8

t

÷ b

y –

9.8

t =

5.8

sec

–2

8.7

–

28

.7

Projectile Motion -2

Name: _____________________

Period: _____________________


6 m/s

Projectile Motion Concepts

Y-direction

Questions

“How high…?” “How much time…” “How long will it be in the air?” - Since the y-direction

is freefall, being pulled down by gravity, these questions are only answered in the y-direction.

Total Speed A projectile’s speed at any point is the vector sum of Vx and Vy at that point (use Pythagorean Theorem).

Only if you are asked for the vertical or horizontal speed do you use one of the components.

Vy =

28.7 m/s

Vx = 41.0 m/s

V = 50 m

/s

35°

Total Speed

Other words:

“Speed”

“Total velocity”

“Initial Velocity”

“Velocity”

Horizontal Speed

Vertical Speed

2 2Total Speed = Vx +Vy

Vyi = Vi =

5 m/s

Vy = V=

0 m/s Since this projectile is

thrown straight up (at

90°), its speed (V) is

equal to the vertical

speed (Vy) at all times.

Since they are at the same height, both balls

will hit the ground at the same time, but

Ball A will go twice as far in that time.

12 m/s B

Greatest Range

v = 35 m/s

60°

vy =

30.3 m/s

vx = 17.5 m/s

D 20°

v = 50 m/s vy =

17.1 m/s

vx = 47.0 m/s C

Object C has the greater initial speed, but object D

will go higher and take longer to get back to the

ground because D has the greater vertical speed.

If projectiles are shot with the same velocity,

but different angles, the range increases the

closer the launch angle gets to 45°. Notice

that 20° and 70° hit at the same point and

both are 25° away from 45°. Also, notice

that 70° goes the highest because it has

v = 3

5 m

/s

60°

vy =

30.3 m/s

vx = 17.5 m/s

20°

v = 50 m/s

vy =

17.1 m/s

vy = 47.0 m/s C

Even though 60° is closer to 45° than 20°, we cannot assume that object D

will have the greatest range, since the balls do not have the same initial

speed. Instead, we must calculate the range as with any ground-to-ground

projectile problem (calculate time in y-direction and range in the

x-direction). This reveals that object C actually has the greater range.

Different Initial Speeds Same Velocity, Different Angles

0

10

20

30

40

50

60

70

80

90

100

0 20 40 60 80 100 120 140 160 180 200

x position (m)

y p

osit

ion

(m

)

45° 35°

20°

65°

70°

Same Initial Speed, Different Angles

D

17.1 = 17.1 9.8

34.2 9.8

3.5 sec

f iv v at

t

t

t

= +

− −

− = −

=

47.0(3.5)

164.5m

DS

T

D ST

D

D

=

=

=

=

30.3 = 30.3 9.8

60.6 9.8

6.18sec

f iv v at

t

t

t

= +

− −

− = −

=

17.5(6.18)

108.2m

DS

T

D ST

D

D

=

=

=

=

How you determine which projectile has the greatest range depends on whether the projectiles

have the same initial velocity.

A

Name: _____________________

Period: _____________________


Projectile Motion Special Situations

All projectile motion problems work the same. First you resolve the initial velocity into Vxi and Vyi. Second, you write

everything you know in the x and y-directions. Third, remembering that ty = tx (times are the same in both directions),

you solve. This, of course, assumes that you know the basics, such as ay = –9.8 m/s2 and ax = 0 m/s2, etc.

Horizontal Launch

2 m/s

4 m

For any horizontally launched object, θ = 0º, Vxi = V and Vyi = 0 m/s.

You should already know:

ay = –9.8m/s2 ax = 0m/s2

∆y = –4 m Vx = 2 m/s

Vyi = 0 m/s

As always, find time in the y-direction:

2

2

2

2

1( ) ( )

2

14 (0 ) ( ( 9.8) )

2

4 4.9

.82

0.9 sec

iy v t at

t t

t

t

t

∆ = +

− = + −

− = −

=

=

Variables:

ay = –9.8m/s2

∆y = –4 m

Vyi = 0 m/s

t = _____

Vyf = not used

( )

2(0 .9)

1 .8 m

DS

T

D or x ST

x

x

=

∆ =

∆ =

∆ =

No acceleration

in the x-direction,

so use the easy

equation!

Maximum Height “Find the maximum height” or “how high?”: these are purely y-direction questions,

so the x-direction can be ignored.

40º

58 m/s

Example 1: A ball is shot 2 m/s horizontally from 4 m up. How far away will it land?

Example 2: A ball is shot 58 m/s at 40º. How high up will it go?

Step 1: Find Vyi.

40º

58 m/s Vyi =

58sin40º

= 37.3 m/s

Write what we know in

the y-direction and solve:

Variables:

ay = –9.8m/s2

∆y = _____

Vyi = 37.3 m/s

Vyf = 0 m/s (at the top)

t = not used

2 2

2

(2 )

0 (37.3 ) (2( 9.8) )

0 1391.29 ( 19.6 )

1391.29 19.9

1391.29

19.9

71 m =

f iV V a y

y

y

y

y

y

= + ∆

= + − ∆

= + − ∆

− = − ∆

−= ∆

−

∆

As with a ground-to-ground example, these two special situations work the same way every time.

More importantly, though, is for you to see the commonality of all projectile motion problems so that you can

solve new problems, when they arise.

How high?

Name: _____________________

Period: _____________________


Assign variables for each of the following situations. Write “?” for any unknown.

y-dir.

ay =

Vyi =

Vyf =

∆y =

t = 3 sec

x-dir.

ax =

Vxi =

Vxf =

∆x =

t =

1. An object is shot 8 m/s horizontally from a 7 m tall cliff. How far away does it land?

2. An object is shot 40 m/s at an angle of 50º. How high does it go?

3. An object is shot 80 m/s at 60º. Calculate range.

y-dir.

ay =

Vyi =

Vyf =

∆y =

t = 6.9 sec

x-dir.

ax =

Vxi =

Vxf =

∆x =

t =

y-dir.

ay =

Vyi =

Vyf =

∆y =

t =

x-dir.

ax =

Vxi =

Vxf =

∆x =

t = 2.5 sec

Projectile Motion Concepts

Name: _____________________

Period: _____________________


y-dir.

ay =

Vyi =

Vyf =

∆y =

t = 2 sec

x-dir.

ax =

Vxi =

Vxf =

∆x =

t =

4. An object is shot 80 m/s at 60º. Find its maximum height.

5. An object is shot 40 m/s at an angle of 50º. How far away does it land?

6. An object is shot 8 m/s horizontally from a 7 m tall cliff. Find its range.

y-dir.

ay =

Vyi =

Vyf =

∆y =

t =

x-dir.

ax =

Vxi =

Vxf =

∆x =

t = 6 sec

y-dir.

ay =

Vyi =

Vyf =

∆y =

t = 3 sec

x-dir.

ax =

Vxi =

Vxf =

∆x =

t =

Name: _____________________

Period: _____________________


1. An object is thrown into the air going 80 m/s at an angle

of 60º. How high does it go?

A. Realizing that in the y-direction projectiles are just

freefall , fill in the y-direction variables.

B. Realizing that in the x-direction, projectiles are at

constant speed, fill in the x-direction variables.

C. In the y-direction, calculate how high the object goes.

2. An object is launched horizontally with a speed of 8 m/s.

A. Since it is launched horizontally, what is the

initial y-direction velocity?

B. What is its initial x-direction velocity?

C. Again, in the y-direction projectiles are just freefall,

fill in the y-direction variables.

D. In the x-direction, projectiles are at constant speed,

fill in the x-direction variables.

F. In the y-direction, calculate how much time it is in

the air before it hits the ground.

G. In the x-direction (at constant speed), what equation

will you use?

H. Calculate how far away it landed in the x-direction,

using the time you just found.

3. An object is shot 40 m/s at an angle of 50º from the

ground. How far away does it land?

A. Fill in the x and y variables for the object.

B. Calculate how long it was in the air, in the

y-direction.

C. In the x-direction (at constant speed), use the time

you just calculated to find how far away it landed.

Projectile Motion Concepts with Diagrams

Name: _____________________

Period: _____________________


4. An object is shot 50 m/s at an angle of 70º.

How high does it go?

A. Use trigonometry to calculate the initial

x and y velocities of the object.

B. Fill in the x and y variables.

C. Calculate how high the object rises.

5. An object is launched 15 m/s horizontally.

A. Fill in the variables for the object.

B. Solve for time in the y-direction.

C. Since the x-direction is constant speed,

solve for ∆x.

6. An object is launched from the ground at a speed of

30 m/s at an angle of 55º. If it lands back on the

ground, calculate how far it went horizontally.

A. Find the initial x and y velocities from the given

speed and direction.

B. Fill in the variables.

C. Calculate time in the y-direction.

D. Calculate ∆x.

Name: _____________________

Period: _____________________


Name: _____________________

Period: _____________________


2. Which has the greater

maximum height?

17. Which has the

greater acceleration?

18. Which has the greater range?

16. Which one has the greater range?


maximum height? 7. Which has the greater

maximum height? 8. Which one hits the ground last?


speed at the top of its path?

11. Which has the lesser

maximum height? 12. Which one hits the ground first?

3. Which one has the greatest range

(lands farthest from the table)?

4. Which one has the least range

(land closest to the table)?

5. Which as the greater

maximum height?

6. Which one has the greatest

acceleration when it leaves the table?

1.Which one hits the ground first?

14. Which one has the smaller

maximum height?


vertical acceleration? 13. Which one has the greater

vertical displacement?

Projectile Motion Comparisons

Name: _____________________

Period: _____________________


Projectile Motion Comparisons -2

28. Which one has the

greatest range?

25. Which is in the air for the

most amount of time?

33. Which one has the greater range?

29. Which as the greater

maximum height?


greatest maximum height?


greater maximum height?


horizontal acceleration? 19. Which has the greater

horizontal velocity? 20. Which one hits the ground first?

24. Which has the

smaller range? 23. Which one has the greatest vertical

acceleration at the top of its path?

22. Which one is in the

air for less time?

27. Which has the

greater range?

30. Which one has the greater

velocity at the top of it’s path?

32. Which one is in the air

for the least amount of time?

Name: _____________________

Period: _____________________


Satellites and Review

Imagine a ball

thrown at the

top of a hill.

Because the

ground falls

as the ball

falls, the ball

goes farther.

For dropped or thrown objects

Vyi = 0 m/s and ay = -9.8 m/s2.

So, in 1 second, it will drop:

21(0 ) ( 9.8)(1)

2

4.9(1) 4.9

y t

y m

∆ = + −

∆ = − = −

1 s

ec =

4.9

m

If you were able to throw the ball

fast enough, the earth would curve

away from the ball as fast as the

ball falls, causing it to be in orbit

around the earth. It would be a

special projectile: a satellite!

Satellites are still falling toward

the earth, but they are going fast

enough to stay in orbit.

The earth’s curvature takes about

8,000 m to falls 4.9 m. To be a satellite,

an object must go 8,000 m in the 1 second

it falls or 8,000 m/sec (8 km/s) or 18,000 mph.

4.9 m

8,000 m (not to scale)

Review

1. Resolve

2. Magnitude

3. Resultant

4. Component

5. Direction

6. Vector

A. “30°” of “35 m at 30°”.

B. “35m” of “35 m at 30°”.

C. How much of the vector is in the

x or y direction.

D. Using trigonometry to finding the x– or

y-components of a vector.

E. The total from start to finish after

adding vectors together.

F. Any physical quantity that requires both

magnitude and direction.

7. Scalar or vector?

A.____A person drives 35 m/s.

B. ____Velocity

C. ____Mass of a person.

D.____acceleration

A C

B D

E

F

G

H

8. A cat climbs 3 m up a tree. Resolve this into its

x and y components.

9. A = – ____

10. B + ___ = 0

11. A + D = ___

12. D – ____ = 2D

13. Which vector/s have +X and –Y components?

14. Which vectors have no x components?

15. Which vectors have equal x and y components?

16. Mathematically, what is A + H—D + 2E?

17. Draw 2E –G +F –C. 18. On the diagram below give two ways to make R.

R A

C

D

B

Name: _____________________

Period: _____________________


19. What is ax?

20. If Vxi = 12 m/s, what is Vxf?

21. So, how do Vxi and Vxf compare?

22. What is ay?

23. If a projectile is launched from the top of

an 9 m tall building, what is ∆y?

24. How do you calculate Vyi?

25. How do you calculate Vxi?

26. In which direction do you calculate time?

27. In which direction is a projectile freefall?

28. In which direction is a projectile in constant motion?

29. On the diagram at the right draw

Vx and Vy at each point on the

projectile’s path.

30. Which one will be in the air the longest?

31. Which one will have the greatest range?

B

A

C

D

E Vx

Vy

A

B

5 m/s

5 m/s 32. Which one goes higher into the air?

33. Which one has the greatest acceleration?

34. Which one has the greatest range?

A B 5 m/s 4 m/s

5 m/s 5 m/s

35. For a projectile shot from the ground to the ground

which angle gives the longest range?

36. Which angle gives a greater range: 20° or 30°?



39. A projectile is shot 25 m/s at an angle of 34°. How high up into the air does the projectile go?

A. Find Vy. C. Solve for “how high”.

B. Write the y-direction variables.

40. A person can swim 2 m/s in still water. The river they are swimming in has a current of 1 m/s.

A. How fast are they swimming relative to the water if they swim with the current?

B. How fast are they swimming relative to the shore if they swim with the current?

C. How fast are they swimming relative to the shore if they swim against the current?

D. If they swim straight across the

river how fast are they swimming

relative to the shore? Vriver

= 1 m/s

Vperson = 1 m/s

Vtotal = ?

Satellites and Review -2

Name: _____________________

Period: _____________________


Name: _____________________

Period: _____________________


Tw

o d

imen

sio

na

l m

oti

on

is

not

as

dif

ficu

lt

as

it m

ay

seem

if

you

un

der

stan

d t

he

log

ic.

Th

is f

low

ch

art

hel

ps

you

ask

th

e

rig

ht

qu

esti

on

s a

nd r

efer

s yo

u t

o t

he

ap

pro

pri

ate

ref

eren

ce m

ate

ria

l

(wo

rksh

eets

) to

un

der

sta

nd

any

two

dim

ensi

on

al

pro

ble

m.

No

te:

Th

is c

ha

rt a

ssu

mes

th

at

all

sit

ua

tio

ns

in w

hic

h V

an

d

a a

re i

n d

iffe

ren

t d

irec

tio

ns

are

pro

ject

ile

mo

tio

n.

Oth

er

po

ssib

ilit

ies

exis

t (a

fo

rce

pu

shin

g a

n o

bje

ct o

ff c

ou

rse,

fo

r

exa

mp

le),

bu

t th

ey a

re l

ess

com

mo

n.

Are

the

unit

s th

e sa

me

for

all

quan

titi

es (

all

m/s

, fo

r ex

am

ple

)?

Is t

he

acce

lera

tio

n (

forc

e) i

n t

he

sam

e d

irec

tio

n a

s th

e vel

oci

ty?

ST

AR

T

No

Yes

Yes

See

wo

rksh

eet:

“A

dd

ing

Vec

tors

”

No

R

eso

lve

vel

oci

ty

into

co

mp

onents

Pla

ce V

x a

nd

Vy i

nto

x a

nd

y d

irec

tio

ns

See

wo

rksh

eet:

“Pro

ject

ile

Mo

tio

n”

Fil

l in

x-d

irec

tio

n

var

iab

les,

rem

em

ber

ing

that

ax =

0 m

/s2

Use

the

kin

em

ati

c

equat

ions

at t

hat

ang

le

(this

is

the

hyp

ote

nu

se)

So

lve

for

un

kno

wn q

uanti

ties

Res

olv

e in

to c

om

po

nents

if

nec

essa

ry (

to f

ind

alt

itud

e [y

],

ho

w f

ast

eas

t [V

x],

etc

)

Is t

he

vec

tor

ver

tica

l o

r ho

rizo

nta

l?

No

Yes

Res

olv

e vec

tor

into

x a

nd

y c

om

po

nen

ts

Lab

el x

or

y

(rem

em

ber

+ o

r -)

Ad

d t

he

all

x’s

and

all

y’s

into

xto

tal a

nd

yto

tal

Fin

d m

ag

nit

ud

e

wit

h P

yth

ago

rean

theo

rem

Fin

d d

irec

tio

n

θ =

tan

-1(y

t/x

t)

Use

sp

ecia

l si

tuat

ions

on

“Fre

e-f

all”

wo

rksh

eet

to

fill

in y

-dir

ecti

on v

aria

ble

s

So

lve

for

any

un

kno

wn q

uanti

ties

y d

irec

tio

n

x d

irec

tio

n

BIG

HIN

T:

Tim

e is

th

e

con

nec

tio

n

bet

wee

n t

he

x a

nd

y

dir

ecti

on

s

So

lve

for

any

un

kno

wn q

uanti

ties

Tw

o D

imen

sio

na

l M

oti

on

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