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

Answer 1

Answer: (D)

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