A particle is released freely from a height H | vedclass

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(B) Let the particle fall from height $H$ and reach a height $h$ from the ground.At this height $h$,the potential energy is $PE = mgh$.The distance fa

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

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(B) Let the particle fall from height $H$ and reach a height $h$ from the ground.At this height $h$,the potential energy is $PE = mgh$.The distance fallen by the particle is $x = H - h$.The kinetic energy at this point is $KE = mgx = mg(H - h)$.According to the problem,$KE = 2(PE)$.Substituting the expressions,we get $mg(H - h) = 2(mgh)$.$H - h = 2h \Rightarrow H = 3h \Rightarrow h = \frac{H}{3}$.Now,to find the speed $v$ at height $h = \frac{H}{3}$,we use the equation of motion $v^2 = u^2 + 2ax$,where $u = 0$ and $x = H - h = H - \frac{H}{3} = \frac{2H}{3}$.$v^2 = 2g(\frac{2H}{3}) = \frac{4gH}{3}$.$v = \sqrt{\frac{4gH}{3}} = 2\sqrt{\frac{gH}{3}}$.Thus,the height is $\frac{H}{3}$ and the speed is $2\sqrt{\frac{gH}{3}}$.

$A$ particle is released freely from a height $H$. At a certain height,its kinetic energy is two times its potential energy. Then,the height and the speed of the particle at that instant are respectively ($g=$ acceleration due to gravity).

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