1. Initial fall

The object is dropped from height $h$. So the first distance covered is $h.$

Just before hitting the ground, its kinetic energy is $K = mgh.$

2. Energy loss at each collision

It loses $50\%$ of its kinetic energy at every hit. So after the first collision, remaining kinetic energy is $K' = \frac{1}{2}mgh.$

This kinetic energy converts into potential energy during the upward motion: $mg h_1 = \frac{1}{2}mgh$ $h_1 = \frac{h}{2}.$

So after first bounce, it rises to height $\frac{h}{2}$ and then falls back the same distance $\frac{h}{2}$.

3. Subsequent bounces

Each time the object hits the ground, its kinetic energy becomes half of what it had just before collision. Hence each rebound height is half of the previous height.

So the heights reached after successive bounces are: $\frac{h}{2},\; \frac{h}{4},\; \frac{h}{8},\; \dots$

4. Total distance covered

Total distance consists of:

• first downward fall: $h$
• then for each bounce, upward and downward travel of twice the rebound height

Therefore, $S = h + 2\left(\frac{h}{2} + \frac{h}{4} + \frac{h}{8} + \cdots \right).$

The series inside brackets is a geometric progression: $\frac{h}{2} + \frac{h}{4} + \frac{h}{8} + \cdots = \frac{\frac{h}{2}}{1-\frac{1}{2}} = h.$

Thus, $S = h + 2(h) = 3h.$

5. Check options

• A: $3h$ ✅
• B: $\infty$ ❌
• C: $\frac{5}{3}h$ ❌
• D: $\frac{8}{3}h$ ❌

Hence the correct answer is $\boxed{3h}.$