PROJECTILES
A projectile is a particle (or an object) which is given an initial velocity and then moves
under the action of its weight alone.
Since we will be ignoring air resistance in our treatment of these projectiles, we can assume
that gravity provides a constant acceleration close to the earth’s surface. This means that the
equations of motion for uniform acceleration can be applied, a being equal to g.
The trajectory is the path taken by the projectile during its motion. This is normally
described as parabolic.
Vectors at right angles to each other are independent.
Vectors at right angles to each other do not have any effect on each other.
An easy example: No matter how hard you push down on an object you will never make it
accelerate sideways.
The vertical and horizontal components of the projectile motion are treated separately
because they represent vectors at right angles to each other.
Vertically the projectile has constant acceleration downwards (due to gravity). If we
choose the upward direction as positive, the acceleration of the projectile in the vertical
direction will be negative because the acceleration always acts downwards.
Horizontally the projectile has constant velocity. Gravity does not act in the
horizontal and we will assume no air resistance so there will be no acceleration (and no
deceleration) in the horizontal direction. Note that if air resistance is considered then there
will be a negative acceleration (deceleration) in the horizontal direction.
There are three cases to consider: (1) vertical projection (2) horizontal projection (3)
oblique projection
CASE 1 – A Particle Projected Vertically Upwards
Consider a particle projected vertically upwards from a point A with velocity u ms-1. The
particle is subjected to a constant acceleration a = -g the vertical component of the particle’s
velocity is u, that is, uy = u. In this case there is no horizontal component of the particles
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initial velocity and there is no horizontal acceleration (gravity does not act horizontally and
we assume no wind) so there will be no horizontal displacement.
Time of Flight (Time to complete journey)
When the projectile returns to the point of projection vertical displacement is zero
y = uyt +
1 2
gt
2
but u y = u , y = 0 and a = -g
1
0 = ut − gt 2
2
1
t u − gt = 0
2
2u
t = 0 OR t =
g
t here is the time of flight of the projectile
Time Taken to Reach Maximum Height
We know that
v y = u y + at
But at maximum height a = -g, u y = u, and v y = 0
0 = u − gt
u
t=
g
where t is the time taken to reach the maximum height
Maximum Height Reached By Particle
From vy2 =uy2 + 2a H
a = -g
uy = u
At the maximum height H MAX →
0 = u2 - 2g HMAX
HMAX = u2
2g
2
Vy = 0
Case 2 – A particle projected horizontally
Consider a particle P projected horizontal from the top of a building h m above ground with
velocity u ms-1. Since g has no components in a horizontal direction, the horizontal
component of the particles’ velocity will remain constant throughout its motion.
We can treat horizontal and vertical motion independently. The particle’s initial vertical
velocity is zero but since g act vertically downward the particle will accelerate at 9.8ms-2.
The particle will take a time t s to reach the ground. We can determine t from the equation:
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- h = ut − gt 2
2
u the initial vertical velocity is zero
t=
2h
g
where t is the time of flight
h is the downward displacement from the starting point so because we choose downward as
the negative direction, both h and g are negative in the equation above.
We can also know how far from the foot of the building the particle hit the ground
(i.e. the horizontal distance or the range, x).
The horizontal distance travelled = velocity time
x=ut
where t is the time of flight as stated in the previous section.
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CASE 3 – A Particle Projected at an Angle to the Horizontal
Consider a particle projected from a point A with velocity u ms-1at an angle to the
horizontal. Since the horizontal and vertical motion is independent of each other we can treat
them separately, using components.
Proof that Projectile Motion is Parabolic
A projectile always travels along a parabolic path. This may be demonstrated by considering
an object projected at angle θ to the horizontal with a speed u.
The vertical component of the initial velocity uy = u sin θ
The horizontal component of the initial velocity ux = u cos θ
Its vertical and horizontal displacement at any time t may be given as
y = (u sin )t − 12 gt 2
x = (u cos )t
Eliminating t in these equations give
y = x tan −
g
x2
2u cos2
2
This is the equation of the path taken by the projectile. It is a quadratic equation (i.e. the
equation of a parabola) of the form y = ax – bx2.
Vertical Motion
The particle is subjected to a constant acceleration a = -g the vertical component of the
particle’s velocity is u sin θ
Time of Flight (Time taken to complete journey)
When the projectile reaches B the vertical displacement is zero
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1 2
gt but u y = u sin , y = 0 and a = -g
2
1
0 = (u sin)t − gt 2
2
1
t u sin − gt = 0
2
2u sin
t = 0 OR t =
g
t here is the time of flight of the projectile
y = uyt +
Time Taken To Reach Maximum Height
We know that
v y = u y + at
v y = u sin − gt
since u y = u sin ,
But at maximum height a = -g and v y = 0
0 = u sin − gt
u sin
t=
g
where t is the time taken to reach the maximum height
NOTE: the time to reach the maximum height is half the time of flight
Maximum Height Reached By Particle
From vy2 =uy2 + 2a H
a = -g
At the maximum height H MAX →
uv = u sin
Vy = 0
0 = u2 sin2 - 2g HMAX
HMAX = u2 sin2
2g
Horizontal Motion
Neglecting air resistance, the horizontal component vx = u cos θ remains constant during the
flight. The horizontal distance travelled AB.
Range = Horizontal velocity time of Flight
= (u cos θ) t
= (u cos θ) 2 u sin θ
g
= u2 (2 sin θ cos )
g
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But remember that
sin 2 = 2 sin θ cos
Range
u2 sin 2θ
g
Now for a given velocity of projection the maximum range occur when sin 2θ is maximum
=
i.e. sin 2θ = 1
2θ = 90
θ = 45
Then the maximum range
RMAX = u2
g
To find direction of motion of a projectile at any time during it’s motion use
tan θ = Vy / Vx
Guidelines for Solving Problems with Projectile Motion
1. To find the time of flight: use s = ut + 12 at 2 , for vertical motion with s = 0.
2. To find time to maximum height: use v = u + at , for vertical motion with v = 0.
3. To find the maximum height: use v 2 = u 2 + 2as , for vertical motion with v = 0.
4. To find the range: find the time of flight, t, and then substitute for t in s = ut for
horizontal motion.
5. To find direction of motion: use tan =
vy
vx
where is the angle the direction of
motion makes with the horizontal, and v y and v x are the vertical and horizontal
components of velocity respectively.
Effect of Air Resistance (Drag) on Projectile Motion
The figure below shows the trajectories for 2 projectiles with the same parameters, 1
projectile experiencing zero air resistance and the other experiencing constant linear air
resistance. The projectile with linear air resistance is the lower curve and shows a
significantly reduced range compared to the “no air resistance” projectile. Not only is the
range reduced when air resistance is present but the shape of the trajectory is also changed.
Specifically, the trajectory with no air resistance is described by a parabola which is
symmetric about the point of maximum height. When air resistance is present the trajectory is
not symmetric and is not parabolic. Instead, it is skewed towards the high-x end of the flight
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because of the diminishing horizontal speed (remember the horizontal speed was constant in
the case of no air resistance).
QUESTIONS
1. (a)
A projectile is fired from the ground, which is flat, with an initial velocity u in a
direction which makes an angle θ with the horizontal.
Show that (i) the maximum height reached by the projectile is (u
sin)2 /2g ,
(ii) the range of the projectile is (u 2 sin cos θ)/g.
Neglect air resistance.
(b)
A cannon ball is fired at 35° to the horizontal with an initial velocity of
200 ms-1. The cannon ball lands in a valley 300 m, below the launch point.
Calculate (i) the maximum height reached by the cannon ball,
(ii) the time of flight of the cannon ball,
(iii)the range of the cannon ball.
Neglect air resistance.
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