An object is dropped from a height h from the ground. Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as `t to oo` is -

Options

A. 2h
B. `oo`
C. `(5)/(3)h`
D. `(8)/(3)`h

Answer

To solve the problem of an object dropped from a height \( h \) that loses 50% of its kinetic energy upon each impact with the ground, we can break down the situation step by step.

### Step-by-Step Solution:

1. **Initial Drop**:
The object is dropped from a height \( h \). When it reaches the ground, it has converted all its potential energy into kinetic energy. The potential energy at height \( h \) is given by:
\[ PE = mgh \]
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity.

2. **First Impact**:
Upon hitting the ground, the object loses 50% of its kinetic energy. Therefore, the kinetic energy just before the impact is:
\[ KE = mgh \]
After the impact, the kinetic energy becomes:
\[ KE' = \frac{1}{2} KE = \frac{1}{2} mgh \]

3. **Rebound Height**:
The kinetic energy after the impact is converted back into potential energy as the object rebounds. The new potential energy at the maximum height \( h_1 \) after the first bounce is:
\[ PE' = KE' = \frac{1}{2} mgh \]
Setting this equal to the potential energy at the new height:
\[ mgh_1 = \frac{1}{2} mgh \implies h_1 = \frac{h}{2} \]

4. **Subsequent Bounces**:
Each time the object hits the ground, it again loses 50% of its kinetic energy. The height it reaches after the second impact \( h_2 \) is:
\[ h_2 = \frac{1}{2} h_1 = \frac{1}{2} \cdot \frac{h}{2} = \frac{h}{4} \]
Continuing this pattern, the heights after each bounce can be described as:
\[ h_n = \frac{h}{2^n} \]

5. **Total Distance Calculation**:
The total distance covered by the object includes the distance falling down and the distance rising up after each bounce. The distances can be summed as follows:
- The first drop: \( h \)
- The first rise: \( h_1 = \frac{h}{2} \)
- The second drop: \( h_1 = \frac{h}{2} \)
- The second rise: \( h_2 = \frac{h}{4} \)
- The third drop: \( h_2 = \frac{h}{4} \)
- The third rise: \( h_3 = \frac{h}{8} \)
- And so on...

The total distance \( D \) can be expressed as:
\[ D = h + 2\left(\frac{h}{2} + \frac{h}{4} + \frac{h}{8} + \ldots\right) \]

6. **Geometric Series**:
The series \( \frac{h}{2} + \frac{h}{4} + \frac{h}{8} + \ldots \) is a geometric series with the first term \( a = \frac{h}{2} \) and a common ratio \( r = \frac{1}{2} \). The sum \( S \) of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{\frac{h}{2}}{1 - \frac{1}{2}} = \frac{\frac{h}{2}}{\frac{1}{2}} = h \]

7. **Final Calculation**:
Substituting back into the total distance:
\[ D = h + 2S = h + 2h = 3h \]

Thus, the total distance covered by the object as \( t \) approaches infinity is \( \boxed{3h} \).