Projectile Motion

Chapter 3.3

Objectives

Recognize examples of projectile motion

Describe the path of a projectile as a parabola

Resolve vectors into their components and apply the kinematic equations to solve problems involving projectile motion

Projectile Motion

How can you know the displacement, velocity and acceleration of a ball at any point in time during its flight?

Use the kinematic equations of course!

Vector Components p.98 (Running vs Jumping)

While jumping, the person is moving in two dimensionsTherefore, the velocity has two components. Vy

Vx

While running, the person is only moving in one dimensionTherefore, the velocity only has one component.V

Definition of Projectile Motion

Objects that are thrown or launched into the air and are subject to gravity are called projectiles

Examples?– Thrown Football, Thrown Basketball, Long

Jumper, etc

Path of a projectile

Neglecting air resistance, the path of a projectile is a parabola

Projectile motion is free fall with an initial horizontal velocity

At the top of the parabola, the velocity is not 0!!!!!!

Vertical and Horizontal Motion

Horizontal Motion Vertical Motion

Velocity = Vx

Displacement = Δx

Velocity = Vy

Displacement = Δy

Because gravity does not act in the horizontal direction, Vx

is always constant!

Gravity acts vertically, therefore a = -9.81 m/s2

Equations for projectiles launched horizontally

Horizontal Motion Vertical Motion

Δx=Vxt

Vx is constant!

a=0

Vy,i=0 (initial velocity in y direction is 0)

ayayayvv iyfy 22022,

2,

atatatvv iyfy 0,,

222, 2

1

2

10

2

1atatattvy iy

Revised Kinematic Eqns for projectiles launched horizontally

Horizontal Motion

Vertical Motion

tvx xyav fy 22

,

atv fy ,

2

2

1aty

Finding the total velocity

Use the pythagorean theorem to find the resultant velocity using the components (Vx and Vy)

Use SOH CAH TOA to find the direction

Vx

Vy

V

Example p. 102 #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? What is the cat’s velocity just before it hits the ground?

What do we know and what are we looking for?

2.2m

Δx= 2.2 m Δy= -1.0m (bc the cat falls down)

1.0

mVx= ?????

What are we looking for??

How do we find Vx?

Equation for horizontal motion:

We have x…so we need t.

How do we find how long it takes for the cat to hit the ground?

Use the vertical motion kinematic equations.

tvx x

Vertical Motion

Δy= -1.0m a=-9.81 m/s^2 What equation should we use?

Rearrange the equation, to solve for t then plug in values.

2

2

1aty

s

smm

a

yt 45.

81.9

)0.1(22

2

Horizontal equation

Rearrange and solve for Vx:

Cat’s Speed is 4.89 m/s

tvx x

s

m

s

m

t

xvx 89.4

45.

2.2

Cliff example

A boulder rolls off of a cliff and lands 6.39 seconds later 68 m from the base of the cliff. – What is the height of the cliff?– What is the initial velocity of the boulder?– What is the velocity of the boulder just as it strikes

the ground?

How high is the cliff?

Δy= ? a=-9.81 m/s2

t = 6.39 s Vx=?

Vy,i = 0 Δx= 68 m

2

2

1aty

ms

maty 200)39.6)(81.9(

2

1

2

1 22

2

The cliff is 200 m high

What is the initial velocity of the boulder?

The boulder rolls off the cliff horizontally

Therefore, we are looking for Vx

tvx x

s

m

s

m

t

xvx 6.10

39.6

68

Important Concepts for Projectiles Launched Horizontally

Horizontal Components Vertical Components

Horizontal Velocity is constant throughout the flight

Horizontal acceleration is 0

Initial vertical velocity is 0 but increases throughout the flight

Vertical acceleration is constant: -9.81 m/s2

Projectiles Launched at An Angle

Projectiles Launched Horizontally

Projectiles Launched at an Angle

•Vx is constant

•Initial Vy is 0

•Vx is constant

•Initial Vy is not 0

Vi

Vx,i

Vy,i

θ

Vi = Vx

Components of Initial Velocity for Projectiles Launched at an angle

Use soh cah toa to find the Vx,i and Vy,i

i

ix

v

v

h

a ,)cos(

i

iy

v

v

h

o ,sin

Vi

Vx,i

Vy,i

θ

cos, iix vv

sin, iiy vv

Revise the kinematic equations again

Horizontal Motion Vertical Motion

2

2

1sin tatvy i

cos, iixx vvv

tvtvx ix cosyavyavv iiyfy 2))sin((2 22

,2

,

atvatvv iiyfy sin,,

Example p. 104 #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?

What do we know?

Δx= 42.0 m

θ= 25°

Vi= 23.0 m/s

Vy at top = 0

Δt=?

Δy=?42.0 m

25°

What can we use to solve the problem?

Find t using the horizontal eqn:

Δx=vxΔt = vicos(θ)t

How to find Δy?– Vy,f = 0 at top of the ball’s path

– What equation should we use?

s

sm

m

v

x

v

xt

ix

0.225)(cos23(

0.42

cos

yavv iyfy 22,

2,

m

sm

sm

a

v

a

vvy iiyfy 8.4

81.92

25sin)23(0

2

)sin(0

2

2

2

22,

2,

Cliff example

A girl throws a tennis ball at an angle of 60°North of East from a height of 2.0 m. The ball’s range is 90 m and it is in flight for 6 seconds.

– What is the initial horizontal velocity of the ball?– What is the initial vertical velocity of the ball?– What is the total initial velocity of the ball?– How high above the initial position does the ball get?– What is the vertical velocity of the ball 2 seconds after it is

thrown?

What is the initial horizontal velocity of the ball?

Δx= 90 m Θ=60° Total time= 6 s Horizontal velocity is

constant: Vx

tvx x

East 156

90

s

m

t

xVx

What is the initial vertical velocity of the ball?

iVx

iVy

,

,tan

North 2698.25)15)(60tan(, s

m

s

mV iy

Vx,i

Vy,i

θ

Vi

What is the total initial velocity of the ball?

2,

2,

2iyixi VVV

Vx,i

Vy,i

θ

Vi

East ofNorth 60at 302615 22 s

mVi

How high above the ground does the ball get?

At the top of the parabola, Vy is 0…so use the revised kinematic equations

Add 2m to get the height above the ground: 36.65 m

yavv iyfy 22,

2,

m

s

ma

vvy iyfy 45.34

81.92

260

22

22,

2,

What is the vertical velocity of the ball 2 seconds after it is thrown?

Vy,i=+26 m/s a= -9.81 m/s2

t = 2 seconds

s

ms

s

m

s

matvv iyfy 4.6)2)(81.9(26

2,,

Important Concepts for Projectiles Launched at an Angle

At the top of the parabola, neither the object’s velocity nor it’s acceleration is 0!!!!!– Only Vy is 0

– Vx is constant throughout the flight

– Horizontal acceleration is always 0– Vertical acceleration is always -9.81 m/s2