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(A) When the object hits the ground,its kinetic energy becomes $50\%$ of its initial value. Let $v$ be the velocity just before impact and $v'$ be the

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$3\,h$

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(A) When the object hits the ground,its kinetic energy becomes $50\%$ of its initial value. Let $v$ be the velocity just before impact and $v'$ be the velocity just after impact.$\frac{1}{2}m(v')^2 = \frac{50}{100} 	imes \frac{1}{2}mv^2 \Rightarrow v' = \frac{v}{\sqrt{2}}$.The coefficient of restitution $e$ is defined as $e = \frac{v'}{v} = \frac{1}{\sqrt{2}}$.The total distance $H$ covered by an object dropped from height $h$ undergoing multiple bounces is given by the formula:$H = h + 2h(e^2) + 2h(e^4) + 2h(e^6) + \dots$This is a geometric series: $H = h + 2h \left( \frac{e^2}{1 - e^2} ight) = h \left( \frac{1 - e^2 + 2e^2}{1 - e^2} ight) = h \left( \frac{1 + e^2}{1 - e^2} ight)$.Substituting $e^2 = \frac{1}{2}$:$H = h \left( \frac{1 + 1/2}{1 - 1/2} ight) = h \left( \frac{3/2}{1/2} ight) = 3h$.

An object is dropped from a height $h$ from the ground. Every time it hits the ground,it loses $50\%$ of its kinetic energy. The total distance covered as $t \to \infty$ is

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