A body is dropped on the ground from a height | vedclass

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(C) Let $V_1$ be the velocity just before hitting the ground and $V_2$ be the velocity just after hitting the ground.Given the ratio of velocities is

Vedclass · Accepted Answer

$375$

Vedclass · Answer

(C) Let $V_1$ be the velocity just before hitting the ground and $V_2$ be the velocity just after hitting the ground.Given the ratio of velocities is $\frac{V_1}{V_2} = 4$,which implies $V_1 = 4V_2$.The kinetic energy just before hitting the ground is $KE_{before} = \frac{1}{2}mV_1^2$.The kinetic energy just after hitting the ground is $KE_{after} = \frac{1}{2}mV_2^2 = \frac{1}{2}m\left(\frac{V_1}{4}ight)^2 = \frac{1}{2}mV_1^2 	imes \frac{1}{16}$.The loss in kinetic energy is $\Delta KE = KE_{before} - KE_{after} = \frac{1}{2}mV_1^2 \left(1 - \frac{1}{16}ight) = \frac{15}{32}mV_1^2$.The percentage loss in kinetic energy is $\frac{\Delta KE}{KE_{before}} 	imes 100 = \left(1 - \frac{1}{16}ight) 	imes 100 = \frac{15}{16} 	imes 100 = \frac{1500}{16} = \frac{375}{4} \%$.Comparing this with $\frac{x}{4} \%$,we get $x = 375$.

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