Question

A particle is released from height $S$ from the surface of the Earth. At a certain height, its kinetic energy is three times of its potential energy. The height from the surface of earth and the speed of the particle at that instant are respectively

• A. $\frac{S}{4},\sqrt{\frac{3gS}{2}}$
• B. $\frac{S}{4},\sqrt{3gS}$
• C. $\frac{S}{4},\frac{\sqrt{gS}}{2}$
• D. $\frac{S}{2},\frac{\sqrt{3gS}}{2}$

Answer 1

The correct option is A $\frac{S}{4},\sqrt{\frac{3gS}{2}}$

Let, the required height of body is $y$.


When body from rest falls through height $(S-y)$. Then under constant acceleration,

$v^2=0^2+2g(S-y)$

$v=\sqrt{2g(S-y)}...\left(1\right)$

When the body is at height $y$ above the ground. Potential energy of body of mass $m$,

$U=mgy$

As per given condition kinetic energy, $K=3U$

$\Rightarrow \frac{1}{2}m(v{)}^2=3\times mg(y)...(2)$

Using equations $\left(1\right)$ and $\left(2\right)$,

$\Rightarrow S-y=3y$

$\therefore y=\frac{S}{4}$

From $\left(1\right)$,

$v=\sqrt{2\times g(S-\frac{S}{4})}=\sqrt{\frac{3gS}{2}}$

Hence, $\left(A\right)$ is the correct answer.