A particle is released from a height H. At a | vedclass

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(D) The total mechanical energy of the particle is conserved,so at any height $h$ from the ground,the sum of potential energy $(PE)$ and kinetic energ

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

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(D) The total mechanical energy of the particle is conserved,so at any height $h$ from the ground,the sum of potential energy $(PE)$ and kinetic energy $(KE)$ is equal to the initial potential energy at height $H$:$PE + KE = m g H$ ... $(i)$Given that at a certain height,the kinetic energy is half of its potential energy:$KE = \frac{1}{2} PE \implies PE = 2 KE$Substituting this into equation $(i)$:$2 KE + KE = m g H$$3 KE = m g H$$KE = \frac{m g H}{3}$Since $PE = m g h$,we have $PE = 2 KE = 2 \left( \frac{m g H}{3} \right) = \frac{2}{3} m g H$.Equating $m g h = \frac{2}{3} m g H$,we get the height $h = \frac{2 H}{3}$.The speed $v$ of the particle at this height can be found using the work-energy theorem or kinematics. Since the particle falls from rest from height $H$,its speed at height $h$ is given by $v = \sqrt{2 g (H - h)}$.Substituting $h = \frac{2 H}{3}$:$v = \sqrt{2 g (H - \frac{2 H}{3})} = \sqrt{2 g (\frac{H}{3})} = \sqrt{\frac{2 g H}{3}}$.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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