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:

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$\frac{S}{4}, \sqrt{\frac{3 gS}{2}}$

$\frac{S}{4}, \frac{3 gS}{2}$

$\frac{S}{4}, \frac{\sqrt{3 gS}}{2}$

$\frac{S}{2}, \frac{\sqrt{3 gs}}{2}$

answer is A.

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Detailed Solution

Conserving energy,
$\begin{array}{l} K + U = mgs \Rightarrow 3 U + U = mgs \\ \Rightarrow 4 mgh = mgs \Rightarrow h = S / 4 \end{array} ∴ K = 3 mg \frac{S}{4} \Rightarrow \frac{1}{2} {mv}^{2} = \frac{3}{4} mgs \Rightarrow v = \sqrt{\frac{3 gs}{2}}$

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