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:

We begin by noting that the body is released from rest at a height $$h$$ above the ground, so on its very first fall it travels a distance $$h$$ straight down to the ground.

On striking the ground the object loses 50 % of its kinetic energy. We recall the basic relation between kinetic energy just before impact and the maximum height attained just after the rebound. If the kinetic energy immediately after the collision is $$K_{\text{after}}$$, then the object will rise to a height $$H$$ given by the conservation of mechanical energy

$$K_{\text{after}} = m g H.$$

Likewise, the kinetic energy just before the collision is $$K_{\text{before}} = m g h_0,$$ where $$h_0$$ is the height from which it has just descended. If a fixed percentage of kinetic energy is retained, then the corresponding percentage of height is retained as well, because in the relation $$K = m g h$$ the mass $$m$$ and the gravitational acceleration $$g$$ remain the same.

Here, only 50 % of the kinetic energy is retained, so 50 % of the height is also retained. Writing this explicitly, if the object falls from a height $$h_n$$ before the $$n^{\text{th}}$$ impact, then after the rebound it will rise to a height

$$h_{n+1} = 0.50 \, h_n = \dfrac{1}{2}\,h_n.$$

Thus the sequence of maximum heights forms a geometric progression:

$$h,\; \dfrac{h}{2},\; \dfrac{h}{4},\; \dfrac{h}{8},\; \dots$$

Let us now add up the total distance travelled. The motion occurs in segments:

1. First fall: a distance $$h$$ downward.
2. First rise: a distance $$\dfrac{h}{2}$$ upward.
3. Second fall: a distance $$\dfrac{h}{2}$$ downward.
4. Second rise: a distance $$\dfrac{h}{4}$$ upward.
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Except for the very first segment, every height in the sequence is traversed twice—once going up and once coming down. Hence we separate the first fall and then double the remaining series of heights:

$$\text{Total distance} = h + 2\!\left(\dfrac{h}{2} + \dfrac{h}{4} + \dfrac{h}{8} + \cdots\right).$$

The expression inside the parenthesis is itself an infinite geometric series with first term $$\dfrac{h}{2}$$ and common ratio $$r = \dfrac{1}{2}.$$ For an infinite geometric series the sum formula is

$$S_{\infty} = \dfrac{\text{first term}}{1 - r}.$$

Substituting the appropriate values, we get

$$S_{\infty} = \dfrac{\dfrac{h}{2}}{1 - \dfrac{1}{2}} = \dfrac{\dfrac{h}{2}}{\dfrac{1}{2}} = h.$$

Putting this result back into the expression for the total distance, we have

$$\text{Total distance} = h + 2 \times h = 3h.$$

Hence, the correct answer is Option A.