Physics 12

Projectile Motion

Clip of the day:

minuteEarth instead today! http://www.youtube.com/watch?v=PRbVISZ3Gc4 http://www.youtube.com/watch?v=uWZwncX_4hk

Projectile Motion

We know that an object (in the absence of air resistance) that is launched at a given angle should follow a parabolic path

We need to break the velocities into x and y components

Review: Projectile Motion – Horizontal Launch

An object that is launched horizontally will have no initial velocity in the y direction so the entire initial velocity will be in the x direction

At this point, we are able to treat the projectile using our two equations of motion

0

00

2

2vtatv

dtvta

td

Recall in 1D:

EQUATIONS OF MOTION

However, in 2D, we need four EOM’s to separate the horizontal and vertical components of motion:

yyy

yyy

y

xxx

xxx

x

vtatv

dtvta

td

vtatv

dtvta

td

0

00

2

0

00

2

2

2

Projectile Motion – Simplified EOMs

Considering there is no acceleration in x and the acceleration in y is g (-9.81m/s2)

yyfy

yfy

xxfx

xfx

dtvtg

d

vtgv

dtvd

vv

00

2

0

00

0

2

Review: Example 1

A cannonball is fired horizontally from the top of a 50.m high cliff with an initial speed of 30.m/s. Ignoring air resistance, determine the following:a. How long it takes to strike the groundb. How far from the base of the cliff it strikes the groundc. How fast it is travelling when it strikes the ground

Start with y position equation (4)

Sub in known information (h=50.m) and solve for time

st

st

sm

mt

ttsm

m

dtvtg

td yyy

2.3

1.10

/81.9

).50(2

002

/81.9.50

2)(

22

22

22

00

2

Now use x position equation (2) Sub in time and known

information (t=3.2s, vox=30.m/s) and solve for dx

msd

ssmsd

dtvtd

x

x

xxx

96)2.3(

0)2.3(/.30)2.3(

)( 00

Finally we will use equations (1) and (3)

Sub in time and solve for velocity

smsv

ssmsv

vtgtv

smv

vtv

y

y

yy

x

xx

/31)2.3(

0)2.3(/81.9)2.3(

)(

/.30

)(

2

0

0

Now, we employ trigonometry and Pythagorean Theorem to solve for the final velocity

o

o

y

x

smv

sm

sm

smv

smsmv

smsv

smv

46,/43

46

/.30

/31tan

/43

)/31()/.30(

/31)2.3(

/.30

1

22

vx

vy

vr

Non-horizontal launch:

Watch video and notice what happens to the velocities http://www.youtube.com/watch?v=hlW6hZkgmkA

Notice: Vx does not change and Vy goes from positive to zero at vertex to negative

Example 2:

A golfer uses a club that launches a golf ball at a 15° angle at a speed of 45m/s. Determine the following:a. The time the golf ball is in the airb. The horizontal distance the ball travelsc. The velocity as it strikes the groundd. The maximum height the ball attains

Start with the position in the y equation

Include the initial velocity in the y term

Solve for when the position in the y is equal to zero

st

smtsm

tsmtsm

tsmtsm

dtvtg

td oyoyy

4.2

)15sin(/45/91.4

)15sin(/45/91.4

0)15sin(/45/91.40

2)(

2

22

22

2

Use the time from the previous question and the position in the x equation

Solve for the range (horizontal distance)

mxsd

ssmsd

dtvtd

x

x

oxoxx

2100.1)4.2(

0)4.2)(15cos(/45)4.2(

)(

Solve for the velocity in the x and y direction

smsv

smssmsv

vtgtv

smv

smv

vtv

y

y

oyy

x

x

oxx

/12)4.2(

)15sin(/45)4.2(/81.9)4.2(

)(

/44

)15cos(/45

)(

2

Use Pythagorean Theorem and Trig to solve for final velocity

o

o

y

x

smv

sm

sm

smv

smsmv

smsv

smv

15,/45

15

/44

/12tan

/45

)/12()/44(

/12)4.2(

/44

1

22

Max height will occur when y velocity is equal to zero. Solve for time and then sub into y position equation

msd

ssmssmsd

dtvtg

td

st

smtsm

vtgtv

y

y

oyoyy

oyy

9.6)2.1(

)2.1)(15sin(/45)2.1(/91.4)2.1(

2)(

2.1

)15sin(/45/81.90

)(

22

2

2

1. A boy kicked a can horizontally from a 6.5 m high rock with a speed of 4.0 m/s. How far from the base of the rock the can land?

2. A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

3. A cannon that launches a cannon ball at a 26° angle at a speed of 33m/s. Determine the following:a. The time the cannon ball is in the airb. The horizontal distance the cannon ball travelsc. The velocity as it strikes the groundd. The maximum height the cannon ball attains

Page 536o Questions 1-8 ANSWERS: 1) 5.9 m 2) 16.5 m/s

3) a) 2.85sec b)86m c)33m/s,334° d) 10m

A FEW MORE TO TRY: