A body is released form the top of a tower of height h meters. It takes t seconds to reach the ground. Where is the ball at the time t/2 sec ?

Options

A. At `(h)/(2)` from the ground
B. At `(h)/(4)` from the ground
C. depends upon mass and volume of the body
D. at `(3h)/(4)` from the ground

Answer

D

Explanation

Let the body after time `(t)/(2)` be at x from the top then ,
`x=(1)/(2)g(t^2)/(4)=(gt^2)/(8)`
`h=(1)/(2)gt^2`
Eliminate t from (i) and (ii), we get `x=(h)/(4)`
Height of the body from the ground `=h-(h)/(4)=(3h)/(4)`