Qno.41 in the exercises of hc verma(i tried a lot but couldn't get it, so i w'll be very grateful if the sol. is explained properly)

here is the solution

boy throws the ball with velocity 9.8m/s, accleration of car is 1m/s²

after the throw ,ball will move in verticle direction to a certain maximum height...

it will take t time to reach to maximum height..

  applying   v =u -gt

 at maximum height velocity will become 0...

  so u =gt     0r

  t=u/g =1sec        (u =9.8 given)

now time taken by ball to reach the maximum height is  t...

time of fall is also t....

 so total time = T=2t=2sec

after two second ball is reached to ground and car has covered some distance during this period...

 distance covered by car is S=ut+at² /2     

                    S=aT² /2               (u is 0 ,a =1 )

                   S=2m ans

Approved