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

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$\frac{S}{2}, \frac{\sqrt{2 g S}}{2}$

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

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

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

answer is D.

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Detailed Solution

Velocity at point C, $v^{2} = 0 + 2 g x Or v^{2} = 2 g x$

Potential energy at C, $P E = m g (S - x)$

At point C, $K E = 3 P E$

$\Rightarrow \frac{1}{2} m \times 2 g x = 3 \times m g (S - x)$

$\Rightarrow x = \frac{3}{4} S$

Height from the ground is $= S - x = S - \frac{3}{4} S = \frac{S}{4}$

Spee of the particle at C, $v^{2} = 2 g x = 2 g \frac{3 S}{4}$

$\Rightarrow v = \sqrt{\frac{3}{2} g S}$

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