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

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$\infty$

$\frac{5}{3} h$

$\frac{8}{3} h$

answer is A.

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Detailed Solution

$h \propto \frac{V^{2}}{2 g}$

$h \propto K . E .$

As K.E. becomes half after every collision. So height will also become half.
So, total distance $= h+2 (\frac{h}{2} + \frac{h}{4} + .....) =h+2h (\frac{\frac{1}{2}}{1 - \frac{1}{2}}) = 3 h$

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