Question

a ball is released from top of a tower of height h meters . it takes t seconds to reach the ground what is the position of the ballin t/3 seconds

Answer 1

The acceleration of the ball will be g.

Initial velocity will be 0.

In T sec the body travels h. 

by applying equations of motion we get

s= ut +1/2gT2

h = 1/2gT2 ------[1]

in T/3 sec h1 = 1/2gT2/9 -------[2]

from [1] and [2] we get h1 =h/9 distance from point of release.

therefore distance from ground is h-h/9 =8h/9.