PHY 1114: Physics I

Lecture 5: Motion in 2D

Fall 2012 Kenny L. Tapp

Quick Question 1A child throws a ball vertically upward at the school

playground. Which one of the following quantities is (are) equal to zero at the highest point of the ball’s trajectory? Assume that at the time of release t = 0, the ball is at y = 0 m.

a) instantaneous velocity

b) displacement

c) instantaneous acceleration

d) average acceleration

e) both instantaneous velocity and instantaneous acceleration3

Quick Question 2• A football is kicked at an angle 25 with respect to the horizontal.

Which one of the following statements best describes the acceleration of the football during this event if air resistance is neglected?

a) The acceleration is zero m/s2 at all times.

b) The acceleration is zero m/s2 when the football has reached the highest point in its trajectory.

c) The acceleration is positive as the football rises, and it is negative as the football falls.

d) The acceleration starts at 9.8 m/s2 and drops to some constant lower value as the ball approaches the ground.

e) The acceleration is 9.8 m/s2 at all times. 4

Quick Question 3A baseball is hit upward and travels along a parabolic arc before it

strikes the ground. Which one of the following statements is necessarily true?

a) The velocity of the ball is a maximum when the ball is at the highest point in the arc.

b) The x-component of the velocity of the ball is the same throughout the ball's flight.

c) The acceleration of the ball decreases as the ball moves upward.

d) The velocity of the ball is zero m/s when the ball is at the highest point in the arc.

e) The acceleration of the ball is zero m/s2 when the ball is at the highest point in the arc. 5

Quick Question 4A ball is launched with an initial velocity as shown. Whichone of the following arrows best represents the direction of theacceleration at point A?

a)

b)

c)

d)

e) The acceleration at point A is zero m/s2.

6

A ball is launched with an initial velocity as shown. Which one of the following arrows best represents the direction of the velocity at point B?

a)

b)

c)

d)

e) The velocity at point B is zero m/s2.

Quick Question 5

7

Components of MotionAn object in motion on a plane can be located using two numbers—the x and y coordinates of its position. Similarly, its velocity can be described using components along the x- and y-axes.

Components of Motion

The velocity components are:

The magnitude of the velocity vector is:

Quick Question 6:• A jetliner is moving at a speed of 245 m/s. The vertical

component of the plane’s velocity is 40.6 m/s. Determine the magnitude of the horizontal component of the plane’s velocity.

10

The horizontal and vertical components of the plane's velocity are related to the speed of the plane by the Pythagorean theorem:

Solving for vh we have:

Quick Question 7:

• If a ball is thrown with a velocity of 25 m/s at an angle of 37 degrees above the horizontal, what is the vertical component of the velocity?

11

Quick Question 8:

• In a football game a kicker attempts a field goal. The ball remains in contact with the kicker’s foot for 0.050 s, during which time it experiences an acceleration of 340 m/s2. The ball is launched at an angle of 51° above the ground. Determine the horizontal and vertical components of the launch velocity.

12

Components of Motion

The equations of motion are:

When solving two-dimensional kinematics problems, each component is treated separately. The time is common to both.

Components of Motion

If the acceleration is not parallel to the velocity, the object will move in a curve:

Quick Question 9:

• A package is dropped from an airplane traveling with a constant horizontal speed of 120 m/s at an altitude of 500 m. What horizontal distance does the package travel before hitting the ground?

15

Projectile Motion

An object projected horizontally has an initial velocity in the horizontal direction, and acceleration (due to gravity) in the vertical direction. The time it takes to reach the ground is the same as if it were simply dropped.

Projectile MotionA projectile launched in an arbitrary direction may have initial velocity components in both the horizontal and vertical directions, but its acceleration is still downward. • Using 2D Motion

principles, why would you travel faster down the tan waterslide compared to the green waterslide? Assume both start from the same height and have the same downward slope.

Quick&Ques)on&10:

Projectile Motion

The range of a projectile is maximum (if there is no air resistance) for a launch angle of 45°.

Projectile MotionWith air resistance, the range is shortened, and the maximum range occurs at an angle less than 45°.

21 12

x = x0 + v0xt + 1/2 axt2 vx = v0x + axt vx2 = v0x

2 + 2axΔx

y = y0 + v0yt + 1/2 ayt2 vy = v0y + ayt vy2 = v0y

2 + 2ayΔx

Position as a function of time

Velocity as a function of time

Velocity as a function of position

2D Equations of Motion(constant acceleration)

13

Zero Launch Angle

x = x0 + v0xt vx = v0x vx2 = v0x

2

y = y0 + v0yt – 1/2 gt2 vy = v0y - gt vy2 = v0y

2 – 2gΔx

Position as a function of time Velocity as a function of time Velocity as a function of position

x = v0t vx = v0 vx2 = v0

2

y = h – 1/2 gt2 vy = - gt vy2 = – 2gΔx

Position as a function of time

Velocity as a function of time

Velocity as a function of position

x0 = 0, y0 = hv0x= v0, v0y= 0

14

Landing site for Zero Launch Angle

• Where does a projectile land if it is launched horizontally with a speed v0 from a height h?

h

x ?

15

Quick Question 11:

A stone is thrown horizontally with an initial speed of 8.0 m/s from the edge of a cliff. A stopwatch measures the stone’s trajectory time from the top of the cliff to the bottom to be 3.4 s. What is the height of the cliff?

16

General launch angle

x0 = 0, y0 = hv0x= v0 cos θv0y= v0 sin θ

18

Projectile Motion: General launch angle• Reference frame chosen

– y is vertical with upward positive• Acceleration components

– ay = -g and ax = 0

• Initial velocity components– vxi = vi cos θ and vyi = vi sin θ

• The velocity components for the projectile at any time t are:* vxf = vxi = vi cos θi = constant

* vyf = vyi + ayt = vyi – g t = vi sin θi – g t

20

General launch angle equations

x = x0 + v0xt vx = v0x vx2 = v0x

2

y = y0 + v0yt – 1/2gt2 vy = v0y - gt vy2 = v0y

2 – 2gΔx

Position as a function of time

Velocity as a function of time

Velocity as a function of position

x0 = 0, y0 = 0v0x= v0 cos θ, v0y= v0 sin θ

x = (v0 cos θ) t vx = v0 cos θ vx2 = v0

2 cos2 θ

y = (v0 sin θ) t - 1/2gt2

y = (v0 sin θ) t - 1/2gt2

vy2 = v0 2sin2 θ –

2gΔx

Position as a function of time

Velocity as a function of time

Velocity as a function of position

using

22

Projectile Motion – Implications

‼ The y-component of the velocity is zero at the maximum height of the trajectory‼ The acceleration stays the same

throughout the trajectory‼ The x-component of the velocity is

constant

24

Range and Maximum Height of a Projectile

• When analyzing projectile motion, two characteristics are of special interest

• The range, R, is the horizontal distance of the projectile

• The maximum height the projectile reaches is h

25

• So, the maximum height of the projectile can be found in terms of the initial velocity vector:

• What are the consequences?– Increase h by making vi or θi bigger or g smaller

Height of a Projectile

26

Range of a Projectile

• Maximum range is TWICE the horizontal position at the time of maximum height.

• In general x = (vi cos θi) t so,

• The range of a projectile can be expressed in terms of the initial velocity vector:

29

Projectile Motion• The maximum range occurs at θi = 45o [sin(2*45o)=1]

The maximum height will be different for two anglesThe times of the flight will be different for two angles

We know... V = Vo - gt--> Vy = Vo sinθ - gtwhere t = T/2, Vy = 0thus, 0 = Vo sinθ - g(T/2)--> T = (2Vo sinθ)/g

• If an object is projected at 50 m/s, 30 degrees above the horizontal, how high and how far will it go?

34

Quick Question 12:

Note:  sin²θ means (sin θ)².

31

Symmetry in Projectile Motion

If launching and landing heights are the same (e. g. y = 0): The time when the projectile lands = 1/2 of the time to

reach its highest point When the projectile lands its speed = the speed of

launching (velocities are different!) The angle of velocity above the horizontal on the way up =

The angle of velocity below the horizontal on the way down33

Non-Symmetric Projectile Motion

• Follow the general rules for projectile motion

or• Break the y-direction

into parts– up and down or – symmetrical back to

initial height and then the rest of the height

– follow the general rules for projectile motion

34

Quick Question 13:• A firefighter a distance 50.0 m from a burning building directs a

stream of water from a fire hose at angle θi = 30° above the horizontal. If the initial speed of the stream is 40.0 m/s, at what height does the water strike the building?

35

Quick Question 13 Solution:

• Water hits the building when

• At this time, the height (y) of the water is

Initial Height

Final Height

Quick&Ques)on&14:• A field goal is attempted when the football is at the center of the

field, 36.58 m from the goalposts. If the kicker gives the ball a velocity of 30 m/s toward the goalposts at an angle of 45° above the horizontal, will the kick be good? (The crossbar of the goalposts is 3.05 m above the ground, and the ball must be higher than the crossbar when it reaches the goalposts for the field goal to be good.)

• Calculate the physics of the rabbit by observing its flight time...

Projectile Motion

• Using the observed rabbit’s flight time, determine the maximum height the rabbit achieves during its flight through the forest.

Maximum HeightInitial Height Final Height

Quick&Ques)on&15:

34

Quick Question 16:• In the gardens of the Bellagio Resort & Casino at Las Vegas, water is

shot up and over a pathway creating an arch. Determine the angle of the jet that shoots the water and the water’s initial velocity to achieve this aqua-feature.

X Y

water missed its mark