Question

A ball is released from the top of tower of
height h meter. If it takes T sec to reach
to ground, where is the ball at time 1/2
second?​

Answer 1

Answer:

This is the answer

Explanation:

We know that

S = ut + 1/2gt2

S = 1/2 gt2     [as initial velocity = u = 0]

S = 1/2gT2

S1  = 1/2g(T/2)2

so S1/S = 1/2gT2/1/2g(T/2)2

    S1 = S * 1/4

    S1 = S/4

so distance from the ground = s - s/4 = 3s/4 or as height is taken h 3h/4