Le Fevre High School
SACE Stage 2 Physics
Projectile Motion 2
1. A cannon on a pirate ship was fired at an angle of 30 °, and the cannon ball came out of the
cannon at 40 ms-1. Assuming we can neglect air resistance and that g=9.8ms -2, calculate:
(a) The vertical and horizontal components of velocity.
v0 horizontal  v0 cos   40  cos 30 0
 34.6ms 1
v0vertical  v0 sin   40  sin 30 0
 20ms 1
(b) How long the cannon ball is in the air?
Half Time of Flight when vt(vertical)  0
vt  v0  at
vt  20  10  t
 t  2s
So the Time of Flight is 4 s
(c) How high it goes?
s vert  v0 t  1 2 at 2
s vert  20  2  1 2  10  2 2
s vert  20m
(d) How far away it lands in the water?
s horiz  v0 t
s horiz  34.6  4
s horiz  139m
(e) What its velocity is after 2.0 seconds?
v2 s  v2 s ( vert )  v2 s ( horiz)
v2 s  0  34.6  34.6ms 1 at 90 0 T
2. A bullet is fired at 30 ° to the horizontal with a speed of 100ms -1 . The acceleration of the bullet
is 10ms -2 vertically downwards.
(a) What is the change in velocity of the bullet after 2 seconds?
Each second, there is a change of velocity of 10ms-1 down.
So after 2 seconds, there is a change of 20ms-1 down.
(b) At what instant after firing will the bullet be travelling horizontally?
Initial components of the velocity of the bullet-
Le Fevre High School
v1( vertical)  v1 sin   100  sin 30 0  100  0.5
 50ms -1
v1( horizontal)  v1 sin   100  cos 30 0  100  0.866
 86.6ms -1
Bullet travelling horizontally when vertical component is zero.
v 2  v1  at
0  50  (10)  t
t  5s
(c) What is the velocity of the bullet 50 seconds after firing?
v50( vertical)  v0( vertical)  at
v50( vertical)  50  (10)  50
v50( vertical)  450ms 1  450ms 1 down
vt  v50( horizontal)  v50( vertical)
vt  86.6  450 
vt  458ms 1 at 169.10 T (Pythagoras' Theory  Trigonometric ratio)
3. A stone is thrown horizontally from the top of a 120m cliff with a speed of 8ms -1.(g=10ms-2)
v0( horizontal)  8ms 1
v0(vertical)  0ms 1
s  120m
(a) How long does it take to strike the sea at the base of the cliff?
Consider the vertical component
s  v1t  1 2 at 2
120  0  1 2  10  t 2
120
5
t  4.9s
t2 
(b) How far from the base of the cliff does it strike the sea?
s  v1( horizontal)  t
s  8  4 .9
s  39.2m
(c) What is its velocity as it strikes the sea?
Strikes the sea after 5s
v5( vertical)  v0( vertical)  at
v5( vertical)  0  (10)  5
v5( vertical)  50ms 1  50ms 1 down
v5  v5( horizontal)  v5( vertical)
v5  8  50  50.6ms 1 170.10 T (Pythagoras' Theory  Trig Ratio)
Le Fevre High School
4. An object is projected horizontally at 30ms -1 from the top of a building which is 40m high. Find:
a  9.8ms 2 down v1  0ms 1 s v  40m t  ?
(a) The time it takes the object to reach the ground.
s v  v1t  1 2 at 2
40  0  1 2  9.8  t 2
40
 8.16
4.9
t  2.86s
t2 
(b) The horizontal distance from the base of the wall to the landing point.
s h  v h  t
s h  30  2.86
s h  85.8m
(c) The time when the object passes a window 25m above the ground.
s  15m
a  9.8ms -2
v1  0ms 1
t  ?
s  v1t  1 2 at 2
15  0  t  1 2  9.8  t 2
15
4.9
t  1.75s
t2 
(d) The objects height above the ground after 2 seconds.
Consider only the vertical component
s  0  1 2  9.8  2 2
s  4.9  4
s  19.6m
5. A cricket ball is hit so that it travels 90m before hitting the ground. If it reaches a maximum
height of 20m calculate the velocity with which the ball left the bat.
Time of flight can be calculated from the ball’s vertical motion. Time to reach maximum
height (at maximum height vvert=0ms-1 )
s vert  vvert t  1 2 at 2
 20  0  1 2  9.8  t 2
20
 4.08
4.9
t  2.02s
Total Time of Flight  2  2.02s  4.04s
t2 
Le Fevre High School
Velocity in Horizontal direction
s
v horiz  horiz
t
90
v horiz 
 22.3ms 1
4.04
Velocity in the vertical direction in reaching the maximum height
s vert  20m
a vert  9.8ms 1
t  2.02s
v 2 vert  0ms 1
v 22vert  v12vert  2as
0  v12vert  2  (9.8)  20
v12vert  392
v1vert  19.8ms 1
Initial velocity can be found by adding the velocity components
v1  v1horiz  v1vert
v12  (22.3) 2  (19.8) 2 (Pythagoras' Theory)
v12  889.33
v1  29.8ms 1 48 0 to the horizontal
6.
(a)
When a projectile lands lower than its launch height, the Time of Flight will
increases by the time it takes to fall the extra height.
Le Fevre High School
(b)
The air resistance will cause an opposing force to the motion of the projectile in the
vertical and horizontal directions. The object will not follow a parabolic path as it would if there
were no air resistance.
300
Projectile ranges
with / without air resistance
250
no
air resistance
height
200
150
100
with
air resistance
50
0
0
200
400
600
800
1000
1200
range
As can be seen the range and Time of Flight are reduced
(c)
When the object is thrown, it leaves at a given angle, the object will then pass the release height at
the same angle. The object thrown at 450 enters the extra ‘space’ at a larger angle dives into the
ground. The object thrown at a less angle enters the extra ‘space’ at a lower angle and can travel
further.