Time taken by the ball to reach ground = 2 sec

when the ball was projected vertically upwards then its horizontal velocity = horizontal velocity of car at the moment when the ball was released = u (say)

so, horizontal distance travelled by the ball in 2 sec = ut = 2u meters ...... (1)

(Remember here there is no component of acceleration for the ball in horizontal direction once it is released)

Distance travelled by the car is given by

s = ut + (1/2)at²

    = 2u + (1/2)*1*(2)²

or s  =2u + 2    ............(2)

From (1) and (2)

The distance at which ball will fall behind the boy on the car is

= (2u + 2 ) - 2u = 2 meters