Question

a body is released from the top of water of height h metres it takes t sec to reach the ground. where is ball at time t/2 sec

Answer 1

Height from which the body is dropped = h meters.
Initial velocity = u = 0
Let the acceleration that the body travels with = a.

Applying the equation of motion: 
  h = u t + 1/2 a t²
     =  1/2 a t²
=>  a = 2h / t²

The distance traveled by the body t/2 sec. after dropping will be:
   s = u (t/2) + 1/2 a (t/2)²
      = 0 + 1/2 * 2h/t² * t²/4  meters.
      = h/4 meters.

So the body/ball is at a height of 3h/4 meters from the ground.

Answer 2

the body is at 3h/ 4 from the ground surface