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(D) Let the height be $h$ and speed be $v$ at the given instant.According to the problem,$K.E. = \frac{1}{2} P.E. \implies P.E. = 2 K.E.$By the law of

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

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(D) Let the height be $h$ and speed be $v$ at the given instant.According to the problem,$K.E. = \frac{1}{2} P.E. \implies P.E. = 2 K.E.$By the law of conservation of mechanical energy,the total energy at any height is constant and equal to the initial potential energy at height $H$:$P.E. + K.E. = m g H$Substituting $P.E. = 2 K.E.$:$2 K.E. + K.E. = m g H \implies 3 K.E. = m g H \implies K.E. = \frac{m g H}{3}$Since $P.E. = m g h$,we have $P.E. = 2 K.E. = 2 \left( \frac{m g H}{3} \right) = \frac{2 m g H}{3}$.Equating $m g h = \frac{2 m g H}{3}$,we get $h = \frac{2 H}{3}$.Now,for the speed $v$:$K.E. = \frac{1}{2} m v^2 = \frac{m g H}{3}$$v^2 = \frac{2 g H}{3} \implies v = \sqrt{\frac{2 g H}{3}}$.Thus,the height is $\frac{2 H}{3}$ and the speed is $\sqrt{\frac{2 g H}{3}}$.

$A$ particle is released from a height $H$. At a certain height,its kinetic energy is half of its potential energy with reference to the surface of the earth. The height and speed of the particle at that instant are respectively:

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