Question

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

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

Answer 1

Answer: (D)

Conserving energy,
$K+U=mgs$
$\Rightarrow 3U+U=mgs$
$\Rightarrow 4mgh=mgs$
$\Rightarrow h=S/4$
$\therefore K=3mg\frac{S}{4}$
$\Rightarrow \frac{1}{2}m{v}^2=\frac{3}{4}mgs$
$\Rightarrow v=\sqrt{\frac{3gs}{2}}$