A ball is dropped from a height of 10, m. If | vedclass

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(C) The initial potential energy of the ball at height $h = 10\, m$ is given by $PE = mgh$.Substituting the values,$PE = m 	imes 10\, m/s^2 	imes 10

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$6$

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(C) The initial potential energy of the ball at height $h = 10\, m$ is given by $PE = mgh$.Substituting the values,$PE = m 	imes 10\, m/s^2 	imes 10\, m = 100m\, J$.After striking the ground,the energy reduces by $40\%$,which means the remaining energy is $100\% - 40\% = 60\%$ of the initial energy.Remaining energy = $0.60 	imes 100m = 60m\, J$.Let the new height reached by the ball be $h'$. The potential energy at this height is $PE' = mgh'$.Equating the remaining energy to the new potential energy: $mgh' = 60m$.$m 	imes 10 	imes h' = 60m$.$h' = 60 / 10 = 6\, m$.Thus,the ball can bounce back to a height of $6\, m$.

$A$ ball is dropped from a height of $10\, m$. If the energy of the ball reduces by $40\%$ after striking the ground,how much high can the ball bounce back (in $,m$)? $(g = 10\, m \,s^{-2})$

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