Two balls are thrown simultaneously, A vertically upwards with a speed of 20 m/s from the ground, and B vertically downwards from a height of 40 m with the same speed along the same line of motion. At what point do the two balls collide? Take `g = 9.8 m//s^(2)`.

Explanation

Let them meet after a time t and at a distance x above the ground. For the first stone, `x= 20t-(1"/"2)9.8 t^(2)` for the second stone `40-x= 20t+(1"/"2)9.8 t^(2)`.
Solving `t= 1s, x= 15.1m`