A body released from the top of tower of height H. it takes ttime to reach the ground where is the body t/2 time after releaseFind the hight at t/2

the acceleration of the ball will be g. Initial velocity will be 0. 

in T sec. body travels h mts.  

 by applying equations of motion we get

                                         s= ut +1/2gT²

                                         h = 1/2gT^{2                                          ------[1]}

in T/2 sec                      h₁ = 1/2gT²/4                                     -------[2]

from [1] and [2] we get h₁ =h/4 distance from point of release.

therefore distance from ground is h-h/4 =3h/4

Regards

Arun (askIITians forum expert)