Question

A body is released from the top of a tower of height$h$. It takes $v = \frac{1}{2}bt^{2} + v_{0}$ sec to reach the ground. Where will be the ball after time $t/2$ sec

• 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 1

Answer: (D)

At $3h/4$ from the ground

Answer 2

Let the body after time $t/2$be at x from the top, then

$x = \frac{1}{2}g\frac{t^{2}}{4} = \frac{gt^{2}}{8}$ …(i)

$h = \frac{1}{2}gt^{2}$ …(ii)

Eliminate t from (i) and (ii), we get $x = \frac{h}{4}$

$\therefore$ Height of the body from the ground $= h - \frac{h}{4} = \frac{3h}{4}$