Question

A body is released from the top of a tower of height $h$. It takes $t\text{sec}$ to reach the ground. Where will be the ball after time $t/2\text{sec}$?

• A. Depends upon mass and volume of the body
• B. At $h/4$ from the ground
• C. At $3h/4$ from the ground
• D. At $h/2$ from the ground

Answer 1

The correct option is C At $3h/4$ from the ground

Given, the body is released from top of the tower, initial velocity, $u=0$

And acceleration, $a=-g$

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

Using $s=ut+\frac{1}{2}a{t}^2$

$x=\frac{1}{2}g\frac{t^2}{4}$

$x=\frac{g{t}^2}{8}\dots \left(i\right)$
And for displacement $h$,
$h=\frac{1}{2}g{t}^2.\dots \left(ii\right)$

Eliminate $t$ from $\left(i\right)$ and $\left(ii\right)$, we get $x=\frac{h}{4}$
$\therefore$ Height of the body from the ground $=h-\frac{h}{4}=\frac{3h}{4}$
Final answer: (d)