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(B) Let the particle be at height $h$ from the ground. The distance fallen by the particle is $(H - h)$.The potential energy at height $h$ is $PE = mg

Vedclass · Accepted Answer

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

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(B) Let the particle be at height $h$ from the ground. The distance fallen by the particle is $(H - h)$.The potential energy at height $h$ is $PE = mgh$.The kinetic energy at height $h$ is $KE = \frac{1}{2}mv^2$. Since the particle falls from rest,$v^2 = 2g(H - h)$,so $KE = mg(H - h)$.According to the problem,$KE = 2PE$.Substituting the expressions: $mg(H - h) = 2(mgh)$.Dividing by $mg$: $H - h = 2h$,which gives $3h = H$,or $h = \frac{H}{3}$.Now,calculate the speed $v$ at this height:$v = \sqrt{2g(H - h)} = \sqrt{2g(H - \frac{H}{3})} = \sqrt{2g(\frac{2H}{3})} = \sqrt{\frac{4gH}{3}} = 2\sqrt{\frac{gH}{3}}$.Thus,the height is $\frac{H}{3}$ and the speed is $2\sqrt{\frac{gH}{3}}$.

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

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