PROJECTILES
1. Horizontal Projection
A projectile is an unpowered object which travels through the air, influenced only by the force of
gravity (we ignore the effect of air resistance in these examples, which can lead to some very
inaccurate answers!) Examples include bullets, cricket balls and custard pies.
We will begin by examining cases where the object is projected horizontally.
Example 1 : A stone is thrown horizontally from the top of a cliff at 10ms-1. The cliff is 80m
high above the sea. Find the time taken for the stone to reach the sea, and its
distance from the base of the cliff when it hits the water.
We consider vertical and horizontal aspects of the motion
10ms -1
separately. This is because there is an acceleration vertically,
namely g, but none horizontally.
Vertically,
80m
s  ut  12 at 2
80  0  12 gt 2
160
g
t  4.04 seconds (2 d.p.)
t2 
Horizontally,
s  ut  12 at 2
 10  4.04
 40.4 m
The stone is airborne for 4.04s, and hits the water 40.4m from the base of the cliff.
Example 2 : Find the distance of the stone in example 1 from its starting position after 2 seconds.
Find also the magnitude and direction of the stone’s velocity at this time.
Vertically,
s  ut  12 at 2
 0  12 g  22
 19.6 m
Horizontally,
s  ut  12 at 2
 10  2  0
 20 m
The distance of the stone from its starting point can be found using Pythagoras’ rule.
distance  19.62  202
 28.0 m (1 d.p.)
We now find the vertical and horizontal components of the stone’s velocity.
Vertically,
v  u  at
 0  8 2
 19.6 ms -1
Horizontally the velocity will always be 10ms-1. Using Pythagoras’ to find the magnitude of the
velocity,
10
v  19.62  102

 22.0 ms (1 d.p.)
-1
v
Using trigonometry to find the direction of the velocity,
19.6
19.6
10
  63.0
After two seconds the stone is travelling at 22.0ms-1, and its direction is 63.0° below the
horizontal.
tan  
Example 3 : A bomb is to be dropped from an aeroplane which is flying steadily at 1000m with a
speed of 200ms-1. How far (horizontally) should the plane be from the target before
it releases the bomb, and how long will the bomb take to hit the target?
For more complicated examples, it helps to set up the initial
information like this.
Notice that since gravity acts vertically, we know there is no
acceleration horizontally, and hence the horizontal velocity
does not change. The vertical acceleration is g, a positive
quantity since it is convenient to define downward motion as
positive. The blank entries in the table indicate unknown
quantities.
To find the time of flight, we consider the vertical motion.
u
v
a
t
s
s  ut  12 at 2
1000  0  12 gt 2
t
2000
g
 14.28 seconds (2 d.p.)
The range is the horizontal displacement at the moment of impact.
s  ut  12 at 2
 200 
2000
0
g
 2855.69 m (2 d.p.)
 2.86 km
Horizontal
200
200
0
Vertical
0
g
1000
2. Projection at Any Angle
Suppose a projectile is launched with velocity u at an angle θ to the
horizontal. The velocity has horizontal component u cos and
vertical component u sin  .
When dealing with these problems, it is again necessary to consider
vertical and horizontal aspects of the motion separately.
u

u sin 
u cos
Example 1 : A stone is thrown at 10ms-1 at an angle of 30° to the horizontal. Find
a) The maximum height obtained by the stone
b) the time for which the stone is airborne.
c) the ‘range’ of the stone (i.e. how far it is from its starting point when it lands).
d) the speed and direction of the stone after 0.3 seconds.
Our initial information can be set out as follows. Note we define upward motion as positive.
Horizontal
Vertical
u
10cos30  5 3
10sin 30  5
v
5 3
0
g
a
t
s
0
a) Max height when vertical v is zero
v2 = u2 + 2as
0 = 52 – 2 × 9.8s
25
s=
19.6
s = 1.28m
b) The stone lands when the vertical displacement is zero.
s  ut  12 at 2
0  5t  12 gt 2
0  12 t (10  gt )
t
10
 1.02 seconds (2 d.p.)
g
We reject the solution t = 0 (which represents the launch of the stone).
c) The range is the horizontal displacement when the stone lands.
s  ut  12 at 2
10
g
 8.84 m (2 d.p.)
 5 3
d) When t = 0.3 seconds, the vertical components of the velocity is given by
v  u  at
 5  0.3 g
 2.06 ms -1
Hence, by Pythagoras,
speed 
5 3 
2
 2.06
v
2
2.06

 8.90 ms-1 (2 d.p.)
5 3
The direction of the motion can be found by trigonometry.
2.06
tan  
5 3
  13.4
The stone is travelling at 13.4° above the horizontal.
Example 2 : A Ming vase is thrown from the top of a tower 15m high, at a speed of 10ms-1 and at
an angle of 75° to the downward vertical. Find
a) the time taken for the vase to smash on the ground.
b) its horizontal range.
c) the speed and direction of motion of the vase just before impact.
75
10ms -1
u
v
a
t
s
Horizontal
10 sin75°
10 sin75°
0
Vertical
10 cos75°
g
15
a) On impact, the vertical displacement is 15 (taking the downward direction to be positive, for
convenience).
s  ut  12 at 2
15  10t cos 75  12 gt 2
1
2
gt 2  10t cos 75  15  0
t
10 cos 75  (10 cos 75) 2  4  12 g  15
g
t  2.03 , t  1.51 seconds (2 d.p.)
This gives us t = 1.51 seconds (rejecting the negative solution).
b) For the range, we require the horizontal displacement when t = 1.51.
s  ut  12 at 2
 10sin 751.51
 14.54 m (2 d.p.)
c) Just before impact, the vertical component of the velocity is given by
v  u  at
 10 cos 75  g 1.51
 17.34 ms -1 (2 d.p.)
Therefore
speed  (10sin 75) 2  17.342
 19.85 ms-1 (2 d.p.)
If θ is the angle between the vase’s direction of motion and the downward
vertical, then
10sin 75
tan  
17.34
  29.1
The vase is travelling at 29.1° to the downward vertical.
10sin 75

v
17.34
Example 3 : Phydeaux the performing dog is to be fired by a cannon across the 200m wide River
Danube. The muzzle velocity is 45ms-1 and the angle of elevation of the barrel is
35°. Does Phydeaux reach the other side, or does he get wet?
Horizontal
Vertical
45 cos35°
45 sin35°
u
45 cos35°
v
g
0
a
t
0
s
Phydeaux lands when his vertical displacement is zero.
s  ut  12 at 2
0  45t sin 35  12 gt 2
0  12 t (90sin 35  gt )
t
90sin 35
g
In this time, the horizontal displacement is
s  ut  12 at 2
90sin 35
0
g
 194.2 m (1 d.p.)
 45cos 35
He gets wet.
Example 4 : Understandably dissatisfied with his manager, Phydeaux gives him the sack, and
considers improving his act in three ways.
a) increasing the velocity of projection u.
b) changing the angle of elevation θ of the barrel.
c) placing the cannon on a platform of height h.
In each case all the other variables are unchanged.
Find the least values of u, θ and h to enable him to reach the other side.
a) The range needs to be 200m, so we have the following initial information.
Horizontal
Vertical
u cos35
u sin35
u
v
u cos35
g
0
a
t
200
0
s
We find an expression for the time of flight from both the horizontal and vertical information,
and compare them.
Horizontally,
s  ut  12 at 2
200  ut cos 35  0
200
t
u cos 35
Vertically,
s  ut  12 at 2
0  ut sin 35  12 gt 2
0  t (u sin 35  12 gt )
t
2u sin 35
g
Comparing the two expressions,
200
2u sin 35

u cos 35
g
200 g  2u 2 sin 35 cos 35
100 g
sin 35 cos 35
u  45.67 ms -1
b) The range needs to be 200m, so we have the following initial information.
u2 
Horizontal
45cos
45cos
0
Vertical
45sin 
u
v
g
a
t
200
0
s
We find an expression for the time of flight from both the horizontal and vertical information,
and compare them. Horizontally,
s  ut  12 at 2
200  45t cos   0
t
200
45cos 
Vertically,
s  ut  12 at 2
0  45t sin   12 gt 2
0  t (45sin   12 gt )
t
90sin 
g
Comparing the two expressions,
200
90sin 

45cos 
g
200 g  4050sin  cos 
200 g  2025  2sin  cos 
200 g  2025sin 2
200 g
sin 2 
2025
2  75.4, 104.6
  37.7, 52.3
Note the clever use of a double-angle formula! The least angle of projection is 37.7°.
Activity 1 : Prove that the range for any speed of projection u is given by
u 2 sin 2
, and prove
g
that an angle of projection of 45° maximises the range.
c) The initial information is...
h
River
u
v
a
t
s
Horizontal
45cos35°
45cos35°
0
Vertical
45sin35°
200
h
g
To find the time of flight, consider the horizontal motion.
s  ut  12 at 2
200  45t cos 35  0
200
t
45cos 35
Using this time in the vertical motion formula,
s  ut  12 at 2
200
 200 
 h  45sin 35
 12 g 

45cos 35
 45cos 35 
h  4.20 m (2 d.p.)
2
Example 5 : A gun with a muzzle velocity of 100ms-1 is fired from the floor of a tunnel which is
4m high. Find the maximum angle of projection possible if the bullet is not to hit the
roof, and the range of the gun with this angle of projection.
For the maximum angle of projection, the bullet must just reach the full height of the tunnel
before starting to fall. This means we require the vertical velocity to be zero when the vertical
displacement is 4.
Horizontal
Vertical
100cos
100sin 
u
0
100cos
v
g
0
a
t
4
s
Vertically, the velocity must be zero at a maximum height of 4m.
v 2  u 2  2as
0  (100sin  ) 2  2  g  4
8g
sin 2  
10000
  5.1
To find the range, we first find the time of flight by considering the vertical motion at the time
s  ut  12 at 2
0  100t sin 5.1  12 gt 2
the bullet lands. 0  t (100sin 5.1  1 gt )
2
t
Horizontally,
200sin 5.1
g
s  ut  12 at 2
200sin 5.1
0
g
 180.0 m (1 d.p.)
 100 cos 5.1
Exercise 6c page 113 questions 1 to 15
3. Trajectory equations
If we consider the launch point of our projectile to be the origin and construct a pair of axes
around this point then by considering the horizontal displacement we can find a formula for the
horizontal displacement (x coordinate) of the projectile in terms of the time from launch, ‘t’. If
we then consider the vertical displacement we can find a formula for the vertical displacement (y
coordinate) of the projectile in terms of t.
Eliminating the time, ‘t’ from these equations will enable us to obtain a formula for the vertical
displacement ‘y’ in terms of the horizontal displacement ‘x’. We call such an equation a
trajectory equation. It is a really useful for:
i)
Checking to see if a projectile can pass through a particular point
ii)
Solving problems when you know the coordinates of some key points
Example 1 : A projectile is launched from the origin with an initial velocity of 20ms-1 at an angle
of 30o to the horizontal.
i)
Write down equations for the position of the projectile after time ‘t’
ii)
Show that the equation of the path is the parabola y = 0.578x – 0.016x2
iii)
Find y when x = 3
iv)
Decide whether the projectile can hit a point 6m above the ground
u
v
a
t
s
i)
Horizontal
20 cos 30
Vertical
20 sin 30
0
g
Horizontally: x = (20cos30)t
Vertically:
y = (20sin30)t – 4.9t2
So x = 17.3t and y = 10t – 4.9t2
ii)
Eliminating t to get a trajectory equation
x
 x 
- 4.9× 

17.3
 17.3 
y = 0.578x – 0.016x2
2
y = 10 ×
iii)
iv)
When x = 3 y = 1.59 using the trajectory equation from part ( ii)
To reach a height of 6 there must be a value of x that gives 6 in the trajectory equation
When y = 6
6 = 0.578x – 0.016x2
0.016x2 – 0.578x + 6 = 0
But the value of b2-4ac is below 0 so this quadratic has no real solutions.
It is therefore impossible for the projectile to reach a height of 6m above the ground.
Example 2 : A batsman hits a cricket ball 0.5 metres above the ground. The ball leaves the bat
with a velocity of 30ms-1, and is caught by a fielder 45 metres away, 1.5 metres
above the ground. Find the two possible angles to the horizontal at which the ball
leaves the bat.
Horizontal
Vertical
30cos
30sin 
u
v
30cos
g
0
a
t
45
1
s
Horizontally,
s  ut  12 at 2
45  30t cos   0
t
3
2cos 
Vertically,
s  ut  12 at 2
1  30t sin   12 gt 2
3
 3 
1  30 
 sin   12 g 

2 cos 
 2 cos  
9g
1  45 tan  
8cos 2 
1  45 tan   11.025sec 2 
2
1  45 tan   11.025(1  tan 2  )
0  11.025 tan 2   45 tan   12.025
45  (45) 2  4 11.025  12.025
2 11.025
  16.0,   75.2
tan  
Notice the quadratic could have has one or no solutions, meaning the fielder is at extreme range
or out of range respectively.
Example 3 : A volleyball player serves from a height of 1 metre, standing 3 metres behind the
net, which is 2 metres high. The player serves at an angle of 35° above the
horizontal. Let y be the height of the ball above the ground when it has travelled
distance x horizontally.
a) Show that
gx 2
1
2u 2 cos 2 35
b) Find the least speed of serve if the ball is to clear the net.
c) Find the greatest speed of serve if the ball is to land in the opposite court, which
extends 5 metres beyond the net.
d) Comment on your answers to b) and c).
y  x tan 35 
a) Horizontally,
s  ut  12 at 2
x  ut cos 35  0
x
t
u cos 35
Vertically,
s  ut  12 at 2
y  ut sin 35  12 gt 2  1
2
 u sin 35
x
x


 12 g 
 1
u cos 35
 u cos 35 
 x tan 35 
gx 2
1
2u 2 cos 2 35
b) Substituting x  3 and y  2 ,
g  32
2  3 tan 35  2
1
2u cos 2 35
9g
1  3 tan 35  2
2u cos 2 35
9g
1  3 tan 35  2
2u cos 2 35
9g
u2 
2
2 cos 35(1  3 tan 35)
u  7.73 ms -1 (2 d.p.)
c) Substituting x  8 and y  0
0  8 tan 35 
g  82
1
2u 2 cos 2 35
64 g
 8 tan 35  1
2u cos 2 35
32 g
u2 
2
2 cos 35(8 tan 35  1)
2
u  8.41 ms -1 (2 d.p.)
d) There is not much margin for error between these two speeds. A steeper angle of projection
would overcome this problem.
Example 4 : A particle is projected with velocity u at an angle θ below the horizontal. At time t,
the particle has travelled a horizontal distance x and a vertical distance y.
a) Prove that
g
x2
2
2u cos 
b) An archer sits on top of a 10m tower, and aims directly at a rabbit 20m from the
base of the tower. If the arrow is fired at 28ms-1, by how much does it fall short
of the rabbit?
a) Horizontally,
y  x tan  
2
s  ut  12 at 2
x  ut cos   0
t
x
u cos 
Vertically downwards,
s  ut  12 at 2
y  ut sin   12 gt 2
x
 x 
 u sin  
 12 g 

u cos 
 u cos  
g
 x tan   2
x2
2
2u cos 
2
b) We have

1
10

20
1
2
,
Substituting y  10, u  28, tan   , cos  
2
5
5

2
10  12 x 
9.8
x2
4
2
2  28  5
9.8 2
x
1254.4
12544  627.2 x  9.8 x 2
10  12 x 
1280  64 x  x 2
x 2  64 x  1280  0
( x  16)( x  80)  0
x  16, x  80
The arrow travels 16m horizontally. It therefore misses the rabbit by 20 − 16 = 4 metres.
Exercise 6E page 123 questions 1 to 10