Projectile Motion

Physics 12

Motion in 2D

We are now going to investigate projectile motion where an object is free to move in both the x and y direction

What is projectile motion?

Any object given an initial thrust and then allowed to soar through the air under the force of gravity only is called a projectile.

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

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

Parts of a Projectile Path

Horizontal Distance = range Height of projectile = altitude, peak

2D Motion of a Horizontal Thrust

Gravity ONLY affects the vertical distance travelled

Gravity is the ONLY force affecting the object (neglect air resistance)

So ax = ay = viy = 0 as there is no initial thrust

given

Projectile Motion - Equations

00

2

0

2)(

)(

dtvta

td

vtatv

)4(2

)(

)3()(

)2(2

)(

)1()(

00

2

0

00

2

0

yyy

y

yyy

xxx

x

xxx

dtvta

td

vtatv

dtvta

td

vtatv

Projectile Motion

yyy

yy

xxx

xx

dtvtg

td

vtgtv

dtvtd

vtv

00

2

0

00

0

2)(

)(

)(

)(

Projectile Motion Problem 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: How long it takes to strike the ground How far from the base of the cliff it

strikes the ground How fast it is travelling when it strikes

the ground

Projectile Motion Problem 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

Projectile Motion Problem 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

Projectile Motion Problem 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

Projectile Motion Problem 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

Example 2

You throw a rock off a 291m high cliff horizontally at 12.8 m/s.

A) If the river below is 68.5 m wide, will the rock make it across the river?

(98.6m so it will make it across) B) With what velocity will the rock hit

the water/ground? (76.6 m/s [80.4’])

Projectile Motion

http://videolectures.net/mit801f99_lewin_lec04/

Page 536-7, questions 1 to 8