A boy standing on a long railroad car throws a ball staight upwards.The car is moving on the horizontal road with an acceleration of 1m/s² and the projection velocity in the vertical direction is 9.8m/s. How far behind the boy will the ball fall on the car?

hi,

till the time , ball is in the hand of bo , it will have same velocity & accl.  of car. & after it is thrown , it will continue in the horizontal velocity with accl. zero , which it has just before leaving the car

let , the velocity of car at the moment of throwing of ball = u

time taken by ball to come back again on the ground  =  2(vertical velo.)/g

                                                                            = 2 sec

during this time distance travelled by ball,  Sb = ut

during this time  distance travelled by the car Sc =  ut + 1/2 a t^2

it shows that ball will fall at a distance   1/2 a t^2 behind the boy

             1/2 a t^2 = 1/2 *1 * (2)^2

                            = 2 m.