Problem : Two projectiles are projected simultaneously from a point on the ground “O” and an elevated position “A” respectively as shown in the figure. If collision occurs at the point of return of two projectiles on the horizontal surface, then find the height of “A” above the ground and the angle at which the projectile "O" at the ground should be projected.

Solution : There is no initial separation between two projectiles in x-direction. For collision to occur, the relative motion in x-direction should be zero. In other words, the component velocities in x-direction should be equal so that two projectiles cover equal horizontal distance at any given time. Hence,

u
O
x
=
u
A
x
u
O
x
=
u
A
x

⇒
u
O
cos
θ
=
u
A
⇒
u
O
cos
θ
=
u
A

⇒
cos
θ
=
u
A
u
O
=
5
10
=
1
2
=
cos
60
0
⇒
cos
θ
=
u
A
u
O
=
5
10
=
1
2
=
cos
60
0

⇒
θ
=
60
0
⇒
θ
=
60
0

We should ensure that collision does occur at the point of return. It means that by the time projectiles travel horizontal distances required, they should also cover vertical distances so that both projectiles are at “C” at the same time. In the nutshell, their times of flight should be equal. For projectile from "O",

T
=
2
u
O
sin
θ
g
T
=
2
u
O
sin
θ
g

For projectile from "A",

T
=
2
H
g
T
=
2
H
g

For projectile from "A",

T
=
2
u
O
sin
θ
g
=
2
H
g
T
=
2
u
O
sin
θ
g
=
2
H
g

Squaring both sides and putting values,

⇒
H
=
4
u
O
2
sin
2
θ
2
g
⇒
H
=
4
u
O
2
sin
2
θ
2
g

⇒
H
=
4
X
10
2
sin
2
60
0
2
X
10
⇒
H
=
4
X
10
2
sin
2
60
0
2
X
10

H
=
20
3
2
2
=
15
m
H
=
20
3
2
2
=
15
m

We have deliberately worked out this problem taking advantage of the fact that projectiles are colliding at the end of their flights and hence their times of flight should be equal. We can, however, proceed to analyze in typical manner, using concept of relative velocity. The initial separation between two projectiles in the vertical direction is “H”. This separation is covered with the component of relative velocity in vertical direction.

⇒
v
O
A
y
=
u
O
y
−
u
A
y
=
u
O
sin
60
0
−
0
=
10
X
3
2
=
5
3
m
/
s
⇒
v
O
A
y
=
u
O
y
−
u
A
y
=
u
O
sin
60
0
−
0
=
10
X
3
2
=
5
3
m
/
s

Now, time of flight of projectile from ground is :

T
=
2
u
O
sin
θ
g
=
2
x
10
x
sin
60
0
10
=
3
T
=
2
u
O
sin
θ
g
=
2
x
10
x
sin
60
0
10
=
3

Hence, the vertical displacement of projectile from "A" before collision is :

⇒
H
=
v
O
A
y
X
T
=
5
3
x
3
=
15
m
/
s
⇒
H
=
v
O
A
y
X
T
=
5
3
x
3
=
15
m
/
s