PROJECTILE MOTION
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1.3 Projectile Motion :
The motion of projectile is known as projectile motion.
It is an example of two dimensional motion with constant acceleration.
Projectile motion is considered as combination of two simultaneous motions in mutually
perpendicular directions which are completely independent from each other Horizontal motion
and Vertical motion.
= +
Parabolic path = vertical motion + horizontal motion.
2. PROJECTILE THROWN AT AN ANGLE WITH HORIZONTAL
2 2
(ucos) +(usin – gt)
R
Consider a projectile thrown with a velocity u making an angle with the horizontal.
Initial velocity u is resolved in components in which horizontal direction is taken as x-axis, vertical
direction as y-axis and point of projection as origin.
ux = u cos uy = u sin
Horizontal direction Vertical direction
(a) Initial velocity ux = u cos Initial velocity uy = u sin
(b) Acceleration ax = 0 Acceleration ay = g
(c) Velocity after time t, vx = u cos Velocity after time t, vy = u sin – gt
Time of flight :
The displacement along vertical direction is zero for the complete flight.
Hence, along vertical direction net displacement = 0
1 2u sin
(u sin ) T – gT2 = 0 T=
2 g
Horizontal range :
2u sin
R = ux .T R = u cos .
g
u2 sin 2
R=
g
Maximum height :
At the highest point of its trajectory vertical component of velocity is zero.
Using v2 = u2 + 2as for vertical direction
u2 sin2
0 = u2 sin2 – 2gH H=
2g
Resultant velocity :
v v x ˆi v y ˆj = u cos î + (u sin – gt) ĵ
Where, | v | = u2 cos2 (u sin gt)2 and tan = vy / vx.
ucos
Also, vcos = ucos v=
cos
General result :
For maximum range = 45º
Rmax = u2/g Hmax = Rmax/2
u2 sin 2 2u sin .ucos 2u x uy
Range can also be expressed as R = = =
g g g
Example 1. A body is projected with a speed of 30 ms–1 at an angle of 30º with the vertical. Find the
maximum height, time of flight and the horizontal range of the motion. [Take g = 10 m/s2]
Solution : Here u = 30 ms–1, Angle of projection, = 90 – 30 = 60º
u2 sin2 302 sin2 60º 900 3 135
Maximum height, H = = = = m
2g 2 10 20 4 4
2u sin 2 30 sin600
Time of flight, T = = = 3 3 sec.
g 10
u2 sin 2 30 30 2 sin 60º cos 60º
Horizontal range = R = = = 45 3 m
g 10
Example 2. A projectile is thrown with a speed of 100 m/s making an angle of 60° with the horizontal. Find
the minimum time after which its inclination with the horizontal is 45° ?
Solution : ux = 100 × cos60° = 50
uy = 100 × sin60° = 50 3
vy = uy + ay t = 50 3 – gt and vx = ux = 50
When angle is 45°,
vy
tan 450 = vy = vx
vx
50 – gt 3 = 50 50 ( 3 1) = gt t = 5 ( 3 1) s
Example 3. A large number of bullets are fired in all directions with the same speed v. What is the
maximum area on the ground on which these bullets will spread ?
Solution : Maximum distance up to which a bullet can be fired is its maximum range, therefore
v2
Rmax =
g
v 4
Maximum area = (Rmax)2 = .
g2
Example 4. The velocity of projection of a projectile is given by : u 5iˆ 10ˆj . Find
(a) Time of flight,
(b) Maximum height,
(c) Range
Solution : We have ux = 5 uy = 10
2u sin 2u y 2 10
(a) Time of flight = = = =2s
g g 10
2
u2 sin2 uy 10 10
(b) Maximum height = = = =5m
2g 2g 2 10
2u sin .ucos 2u x uy 2 10 5
(c) Range = = = = 10 m
g g 10
Example 5. A particle is projected at an angle of 30° w.r.t. horizontal with speed 20 m/s :
(i) Find the position vector of the particle after 1s.
(ii) Find the angle between velocity vector and position vector at t = 1s.
Solution :
3
(i) x = u cos t = 20 × ×t = 10 3 m
2
1 1
y = u sin t – × 10 × t2 = 20 × × (1) – 5 (1)2 = 5m
2 2
2
Position vector, r = 10 3 ˆi 5ˆj , | r |
10 3 52
(ii) vx = 10 3 î
vy = uy + ayt = 10 – g t = 0
v = 10 3 î , | v | = 10 3
v r (10 3î) (10 3î 5 ĵ) = 300
v . r = | v | | r | cos
vr 300 3
cos = = = cos 1 2
|v||r | 10 3 325 13
3. EQUATION OF TRAJECTORY
The path followed by a projectile during its motion is called its Trajectory.
Equation of trajectory is the relation between x & y coordinate of the particle.
x
y = x tan 1
R
Example 1. Find the value of in the diagram given below so that the projectile can hit the target.
gx 2 (1 tan2 ) 5 (20)2
Solution. y = x tan – 10 = 20 tan – (1 + tan2)
2u2 (20)2
2 = 4 tan – (1 + tan2)
tan2 – 4 tan + 3 = 0
(tan – 3) (tan – 1) = 0 tan = 3, 1 = 45°, tan–1(3)
Example 2. A ball is thrown from ground level so as to just clear a wall 4 m high at a distance of 4 m and
falls at a distance of 14 m from the wall. Find the magnitude and direction of initial velocity of
the ball figure is given below.
Solution.
P
The ball passes through the point P(4, 4). Also range = 4 + 14 = 18 m.
x
The trajectory of the ball is, y = x tan (1 – )
R
Now x = 4m, y = 4m and R = 18 m
4 7 9 9
4 = 4 tan 1 = 4 tan . or tan = = tan–1
18 9 7 7
2u2 sincos 2 9 7
And R = or 18 = × u2 × × u= 182
g 9.8 130 130
4. PROJECTILE THROWN PARALLEL FROM HEIGHT
(NOT IN SYLLABUS)
Consider a projectile thrown from point O at some height h from the ground with a velocity u.
Horizontal direction Vertical direction
(i) Initial velocity ux = u Initial velocity uy = 0
(ii) Acceleration ax = 0 Acceleration ay = g (downward)
4.1 Time of flight : 4.2 Horizontal range :
From equation of motion .
1
S = ut + at2, along vertical direction, we get 2h
2 R = ux . t R =u
1 g
– h = uyt + (–g)t2
2
1
h = gt2
2
2 g
h
t=
4.3 Velocity at a general point P(x, y) :
v = u2x u2y
Horizontal velocity after time t
vx = u
Vertical velocity after time t
vy = 0 + (–g)t = –gt = gt (downward)
v = u2 g2 t 2 and tan = vy/vx
4.4 Velocity with which the projectile hits the ground :
Vx = u
Vy2 = 02 – 2g(–h)
Vy = 2gh u
2
g
h
2
V= Vx2 Vy2 V=
Solved Example
Example 1. A projectile is fired horizontally with a speed of 98 ms–1 from the top of a hill 490 m high. Find
(i) the time taken to reach the ground
(ii) the distance of the target from the hill and
(iii) the velocity with which the projectile hits the ground. (take g = 9.8 m/s2)
Solution :
(i) The projectile is fired from the top O of a hill with
speed
u = 98 ms–1 along the horizontal as shown as OX.
It reaches the target P at vertical depth
OA, in the coordinate system as shown,
OA = y = 490 m
As, y = 1/2 gt2
1
490 = × 9.8 t2
2
or t = 100 = 10 s.
(ii) Distance of the target from the hill is given by, AP = x = Horizontal velocity × time = 98 × 10 = 980 m.
(iii) The horizontal and vertical components of velocity v of the projectile at point P are
vx = u = 98 ms–1
vy = uy + gt = 0 + 9.8 × 10 = 98 ms–1
V = v 2 v 2 = 982 982 = 98 2 ms–1
x y
Now if the resultant velocity v makes an angle with the horizontal, then
vy 98
tan = = =1 = 45º
vx 98
