Q 11-05-197JEE MainJEE Main 2017 (8 Apr)Medium
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:
Answer: (A) $3h$
After each impact the kinetic energy, and therefore the height reached, is halved: $h/2,\ h/4,\ h/8,\dots$ Each rebound height is travelled up and down:
$$d = h + 2\left(\frac h2 + \frac h4 + \dots\right) = h + 2\cdot\frac{h/2}{1 - 1/2} = h + 2h = 3h$$
Solution by Sreeraj P, M.Sc Physics