A body is released from the top of a tower of height h. It takes t sec to reach the ground. Where will be the ball after time `(t)/(2)` sec?

Options

A. At `(h//4)` m from the ground
B. At `(h//2) m` from the ground
C. At `(3h//4) m` from the ground
D. Depends upon the mass and volume of the ball

Answer

C

Explanation

`T = sqrt((2h)/(g)), h' = (1)/(2) xx g xx ((T)/(2))^(2)`