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

\frac{H}{3}, \sqrt{\frac{2 g H}{3}}

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\frac{H}{3}, 2 \sqrt{\frac{g H}{3}}

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\frac{2 H}{3}, \sqrt{\frac{2 g H}{3}}

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\frac{H}{3}, \sqrt{2 g H}

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Solution

The correct option is B $\frac{H}{3}, 2 \sqrt{\frac{g H}{3}}$
Using law of conservation of energy,
$K_{i} + U_{i} = K_{f} + U_{f}$
let initial position be when particle is dropped from height $H$
$U_{i} = m g H$
$K_{i} = 0$
Given $K_{f} = 2 U_{f}$
$K_{i} + U_{i} = K_{f} + U_{f}$
$m g H = 3 U_{f}$

$U_{f} = \frac{1}{3} m g H$ and $K_{f} = \frac{2}{3} m g H$
Now using $K_{f} = \frac{m v^{2}}{2}$
and speed of particle will be
$v = \sqrt{2 g (\frac{2 H}{3})} = 2 \sqrt{\frac{g H}{3}}$
Now using $U_{f} = m g h$
$h = \frac{H}{3}$

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