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

 
v (t )  at  v0
2



at
d (t ) 
 v0t  d 0
2



v x (t )  a x t  v0 x (1)
 2



axt
 v0 x t  d 0 x (2)
d x (t ) 
2



v y (t )  a y t  v0 y (3)
 2


a yt

 v0 y t  d 0 y ( 4 )
d y (t ) 
2
Projectile Motion


vx (t )  v0 x



d x (t )  v0 xt  d 0 x
 

v y (t )  gt  v0 y
2



gt
 v0 y t  d 0 y
d y (t ) 
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
2



gt
d y (t ) 
 v0 y t  d 0 y
2
 9.81m / s 2t 2
 50.m 
 0t  0
2
2(50.m)
2
t 
 9.81m / s 2
t 2  10.1s 2
t  3.2 s
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



d x (t )  v0 x t  d 0 x

d x (3.2 s )  30.m / s (3.2 s )  0

d x (3.2 s )  96m
Projectile Motion Problem
 Finally we
will use
equations (1)
and (3)
 Sub in time
and solve for
velocity


v x (t )  v0 x

v x  30.m / s

 
v y (t )  gt  v0 y

v y (3.2 s )  9.81m / s 2 (3.2s )  0

v y (3.2 s )  31m / s
Projectile Motion Problem

 Now, we employ v x  30.m / s
trigonometry and vy (3.2 s )  31m / s
Pythagorean
2
2
v

(
30
.
m
/
s
)

(

31
m
/
s
)
Theorem to solve
for the final
v  43m / s
velocity
vr
31m / s
  tan
30.m / s
o
  46

v  43m / s,46o
1
vx
vy
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_le
win_lec04/
 Page 536-7, questions 1 to 8