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 → ∞ is:

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TestHub · Accepted Answer

Since the object is dropped from height, thus its potential energy is equal to its kinetic energy.

As per question, kinetic energy after hitting the ground is $\frac{1}{2} m v^{' 2} = \frac{1}{2} \times \frac{1}{2} m v^{2}$

$v^{'} = \frac{v}{\sqrt{2}}$

$v = e u$

$e = \frac{1}{\sqrt{2}}$

$H = h (\frac{1 + e^{2}}{1 - e^{2}})$

The total distance covered $= h (\frac{1 + \frac{1}{2}}{1 - \frac{1}{2}}) = 3 h$

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