Consider the situation as shown in the figure.

Two balls are thrown simultaneously with same speed `10 m//s`, one horizontally and another at angle `60^(@)` in downward direction. After sometime balls cllide in mid-air, find distance `d`.

Explanation

Taking `B` as origin, let the balls collide after time `t` and at point `P(x, y)`.

Ball (1): `x = 10 t` (i)
`y = 20 - (1)/(2)g t^(2)` (ii)
Ball (2): `d - x = 10 cos 60^(@) t`
(iii) `y = 30 - {10 sin 60^(@) t + (1)/(2)g t^(2)}`
(v) Equating (ii) and (iv), we get
`20 - (1)/(2)g t^(2) = 30 - {10 sin 60^(@) t + (1)/(2)g t^(2)}`
`20 = 30 - (10sqrt(3))/(2)t rArr t = (2)/(sqrt(3))s`
Adding (i) and (iii), we have
`d = 10t + 10 cos 60^(@) t`
`d = 15t = 15 xx (2)/(sqrt(3)) = (30)/(sqrt(3)) = 10sqrt(3) m`