Question

A ball released from certain height h reaches the ground in time T. Where it will
5T
be from the point of release at time ? (From the top]
6​

Answer 1

The ball displacement when it reaches the ground in time T is h .

Hence ,

s = h , t = T , a = g , u = 0

Here s is displacement of the body, in this case ball , in time t moving with acceleration which in this case is acceleration due to gravity g , u is initial velocity given to the body at the start of the displacement, which is 0 in this case gotten the ball is dropped and not thrown.

NOTE :- All the measurements are made write a frame of reference in which downward direction is measured positive.

According to equation of motion

s = ut + (1/2)at²

h = 0 + (1/2)gT²

T² = 2h/g

Let us apply the same equation of motion for displacement after time T /2

s = ut + (1/2)at²

s = 0 + (1/2)g(T/2)²

s = (1/2)(g/4)T²

Substituting the value of T²

s = (1/2)(g/4)(2h/g)

s = h/4

Hence the ball falls distance of h /4 in time T/2 , i.e. the ball will be at a height of 3h / 4 or 0.75 h after time T/2.

hope it helps