PROJECTILES
THE CONCEPTS OF PROJECTILES
Let us consider a boy who
releases a piece of stone from his catapult against a bird on a tree branch,
the stone will travel in parabolic path towards the bird. Likewise if we throw
a tennis ball against a wall, the path of the ball towards the wall is a
parabola. On hitting the wall, the ball returns to the ground also along a
parabolic path. The same type of curve is seen when a ball is projected
horizontally from the top of a building. The stone or ball been projected is
known as PROJECTILE.
EXAMPLES OF PROJECTILE MOTION
(i)
A thrown rubber
ball re – bouncing from a wall.
(ii)
An athlete doing
the high jump.
(iii)
A stone released
from a catapult.
(iv)
A bullet fired
from a gun.
MOTION OF A PROJECTILE
Let us consider the simple
case of a stone thrown horizontally with an initial velocity of u from the top
of a high wall of height h. The stone is subject to two independent motions;
(i)
A horizontal
motion: once the stone is free and before it hits the ground and;
(ii)
A vertical motion:
when the stone is initially thrown forward, the vertical pull of gravity will
accelerate it downwards at a rate of about 9.8ms-1.
Any
projectile in flight is doing two things at once; flying horizontally with a
constant speed and moving up and down with an acceleration of g.
Example
1: A tennis ball is projected horizontally from the top of a vertical cliff 50m
high with a velocity of 10ms-1. Find:
(a) The time to reach the ground.
(b) The vertical component of the velocity when the ball
hits the ground.
(c) The distance from the foot of the cliff where the ball
hits the ground.
SOLUTION
(a) The ball falls vertically under the influence of
gravity. Its vertical motion is independent of the horizontal motion.
Initial
vertical velocity = 0, acceleration a = g, h = 50m
Recall
from equations of motion, S = ut + ½at2
h
= ut + ½gt2
50
= 0 + ½ x 10 x t2
50
= 5 t2
t2 = 10
t = √10
t = 3.16s.
(b) Let v be the vertical component of the velocity when
the stone hits the ground.
From equations of motion, v =
u + at.
v = u + gt
v = 0 + 10 x 3.16
v = 31.6ms-1.
(c) Let s be the distance from the foot of the cliff where
the ball hits the ground. Initial horizontal velocity is 10ms-1 and
not affected by vertical acceleration therefore, a = 0.
From equations of motion, S =
ut + ½at2
S = ut + 0 (since a = 0)
S = 10 x 3.16
S = 31.6m.
PATH OF A
PROJECTILE
TIME TO REACH THE MAXIMUM
HEIGHT
From the figure above the vertical component is given as Ux.
Therefore, the initial velocity u of the projectile is given as;
U = USinθ …………………………… (i)
Recall from equations of motion, V = u + gt.
Since the body is thrown up, so the above expression will be
V = u – gt …………………………….(ii)
Substitute (i) into (ii)
V = USinθ – gt …………………….(iii)
At maximum height, final velocity V = 0
0 = USinθ – gt
gt = USinθ
t = USinθ 
g ……………………..(iv)
Equation (iv) is the time to reach the maximum height.
TIME TO FLIGHT
This is the time to complete the projectile path which is 2 x (time to
reach the maximum height).
T = 2 x USinθ
g
T = USinθ …………………………... (v)
g
Equation (v) is the time to flight.
MAXIMUM HEIGHT
From equation of motion, v2 = u2 + 2as
Since the body is thrown up, so the above expression will be
v2 = u2 – 2gH
But at maximum height, v = 0
0 = u2 – 2gH
2gH = u2
Since U = USinθ
Therefore, 2gH = (USinθ)
2
H = U2Sin2θ
2g
The above expression is the value of the maximum height of a
projectile.
THE RANGE (R)
The range of a projectile is given as:
the horizontal component x time to flight.
That is, UCosθ x T
Since T = 2USinθ
g
R = UCosθ x 2USinθ
g
R = UCosθ x 2USinθ
g
R = U22CosθSinθ
g
but from trigonometric, 2CosθSinθ = Sin2θ
R = U2Sin2θ
g
The above expression is the value of the Range of a projectile.
Example: A projectile is fired with an initial velocity of 100ms-1
at an angle of 30 to the horizontal. Calculate:
a. The time of flight.
b. The maximum height.
c. The range.
Solution
a. T = USinθ = 2 x 100 x 0.5
= 10s.
g 10
H
= U2Sin2θ = (100) 2 Sin230 = 104 x 0.25 = 125m.
2g 2x10 20
c. 
R = U2Sin2θ = (100) 2 Sin 2(30)
= 104 x 0.8660 = 866 m.
R = U2Sin2θ = (100) 2 Sin 2(30)
= 104 x 0.8660 = 866 m. 
g 10 10
great work. simplified and understandable. thanks
ReplyDeleteSo easy I couldn't believe I just got the answer same as my teachers
ReplyDeleteWhy cant we use
ReplyDeleteV=u +at to calculate the time taken to reach the ground in example 1a
Thanks. You can't use V = u + at because the angle of projection is given.
Delete