A ball of mass 10 text{ g} is allowed to fall | vedclass

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(C) Initial potential energy of the ball at height $H = 10 	ext{ m}$ is $E_i = mgH$.When the ball hits the ground, this energy is converted into kine

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Initial potential energy of the ball at height $H = 10 	ext{ m}$ is $E_i = mgH$.When the ball hits the ground, this energy is converted into kinetic energy.After the collision, $50\%$ of the energy is lost, so the remaining energy is $E_f = 0.5 	imes E_i = 0.5 	imes mgH$.The ball will rise to a new height $h$ such that its potential energy at that height equals the remaining energy:$mgh = 0.5 	imes mgH$$h = 0.5 	imes H$Given $H = 10 	ext{ m}$, we have:$h = 0.5 	imes 10 	ext{ m} = 5 	ext{ m}$.Therefore, the height reached by the ball is $5 	ext{ m}$.

$A$ ball of mass $10 \text{ g}$ is allowed to fall from a height of $10 \text{ m}$. After the collision with the ground, if $50\%$ of its energy is lost, then the height reached by the ball is: (in $\text{ m}$)

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