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Step-by-Step Solution
Step 1: Initial Drop
The object is initially dropped from a height $h$, so the first distance traveled by the object before it hits the ground is $h$.
Step 2: Velocity Before Impact
Just before hitting the ground the first time, all of the object’s initial potential energy $mgh$ is converted into kinetic energy. Hence, if $v$ is the velocity just before impact:
$$
\frac{1}{2} m v^2 = mgh
\quad \Longrightarrow \quad
v = \sqrt{2gh}.
$$
Step 3: Loss of Kinetic Energy on Impact
Every time the object hits the ground, it loses 50% of its kinetic energy. This means the new kinetic energy $k'$ after collision is half of the previous kinetic energy $k$:
$$
k' = \frac{1}{2} k.
$$
Since $k = \frac{1}{2} m v^2$, the velocity after rebound, say $v_1$, satisfies
$$
\frac{1}{2} m v_1^2 = \frac{1}{2} \left( \frac{1}{2} m v^2 \right)
\quad \Longrightarrow \quad
v_1^2 = \frac{v^2}{2}
\quad \Longrightarrow \quad
v_1 = \frac{v}{\sqrt{2}}.
$$
Step 4: Height After Rebound
The height $h'$ to which the object rises after the bounce is given by
$$
h' = \frac{v_1^2}{2g} = \frac{1}{2g}\left(\frac{v^2}{2}\right) = \frac{v^2}{4g}.
$$
But originally $v^2 = 2gh$, so
$$
h' = \frac{2gh}{4g} = \frac{h}{2}.
$$
Step 5: Distance Covered in Each Complete Bounce
After the first drop of $h$, the object bounces up to a height of $\frac{h}{2}$ and then falls back the same distance $\frac{h}{2}$. Hence, each bounce (going up and coming down) covers a total additional distance:
$$
\frac{h}{2} + \frac{h}{2} = h.
$$
Then in the next bounce, the height becomes $\frac{h}{4}$ (half of $\frac{h}{2}$ again), and so on.
Step 6: Summing Up the Infinite Series
The total distance covered consists of:
The first descent of $h$
An infinite geometric series of bounces, each contributing $h$, $\frac{h}{2}$, $\frac{h}{4}$, etc.
However, note that the “double travel” of each bounce (up and down) can be collectively viewed as an infinite sequence with a ratio of $\tfrac{1}{2}$. More explicitly:
$$
\text{Total distance} = h + \bigl(h + \tfrac{h}{2} + \tfrac{h}{4} + \dots \bigr).
$$
The sum of the infinite series $h + \tfrac{h}{2} + \tfrac{h}{4} + \dots$ is
$$
\frac{h}{1 - \frac{1}{2}} = 2h.
$$
So the total distance becomes
$$
h + 2h = 3h.
$$
Final Answer
$\boxed{3h}$
