Projectile Motion

Horizontally Launched ProjectilesProjectiles which have NO upward trajectory and NO

initial VERTICAL velocity.

Y-DirectionFree Fall – Dropping Problem

acceleration = -9.8m/s2

X-DirectionNo air resistance = no accelerationCONSTANTVV ixi

tdVX

RANGEd

smViy /0

yAll acceleration equations apply

The amount of time to fall ∆y = the amount of time to travel ∆d

Class Demo: Arrow Launcher

Example: In movies, you often see people jump from one building roof to

another. If a person were to jump from a roof with a speed of 2.5m/s and the roof he wants to land on is 3.5m away and is 4.5m lower than the roof is jumping from, will he make it to the other roof?

Projectiles Launched At An AngleSince the projectile was launched at a

angle, the velocity MUST be broken into components!!!

vi

vix

viy

q

cos

cos

iiX

i

iX

VVVV

X-Direction

Y-Direction

sin

sin

iiy

i

iy

VVVV

CONSTANTVix

Free Fall – Up and Down Problem

acceleration = -9.8m/s2

tVd i cos2

22

21sin

2)sin(

sin

tgtVy

ygVV

tgVV

i

ify

ify

Projectiles Launched At An Angle

Horizontal Velocity is constant

Vertical Velocity decreases on the way upward

Vertical Velocity increases on the way down,

NO Vertical Velocity at the top of the trajectory.

Component Magnitude DirectionHorizontal Constant ConstantVertical Decreases up, 0

@ top, Increases down

Changes

Vertically Launched ProjectilesThere are several things you must

consider when doing these types of projectiles besides using components.

1. If it begins and ends at ground level, the “y” displacement is ZERO: ∆y = 0

2. The amount of time to go up to ymax is the same time to come down.

3. The velocity Vi that it was launched (speed and direction) is the same when it lands Vf

4. The maximum range is at 45o

5. Complimentary angles give the same range, BUT NOT the same flight time

A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees.

(a) How long is the ball in the air?(b) How far away does it land?(c) How high does it travel?

tgVV ify sin

a) Vfy=0 (top), ∆t=tup

sec2.3

2

sec6.1/8.9

)53sin/20(

sin

2

total

uptotal

up

up

up

t

tt

tsm

mt

gVt

b) ∆dt =?, need ttotal

mdsmdtVd i

5.38sec2.3)53cos()/20(

cos

c) ∆ymax = ? Use Vfy = 0m/s and tup

mysm

smy

gVy

ygVV

i

ify

13)/8.9(2))53(sin/20(

2)sin(

2)sin(

max

2

2

max

2

max

22