A man standing on a long rail road car throws a ball vertically upward with speed `20 m//s`. The car starts from rest and moves with acceleration `2 m//s^(2)`. How far behind the man will the ball fall on the car ?

Explanation

The ball will return to the point of projection `O` after time
`y = v_(0) t - (1)/(2) g t^(2)`
`0 = 20 t - (1)/(2) xx 10 t^(2) rArr t = 4 s`
The distance travelled by the man or the car
`s = u t + (1)/(2) a t^(2)`
`d = 0 + (1)/(2) xx 2 xx 4^(2) = 16 m`
The ball will fall a distance `16 m` behind the man. The amswer will be same if car is having some initial velocity.
Note: The ball will far at distace `(1)/(2) a t^(2)` behind/ahead of man if the car is moving under constant acceleration/deceleration `a`.