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

Explanation

To solve the problem step by step, we need to analyze the motion of the object dropped from height \( h \) and how its kinetic energy changes with each bounce.

### Step-by-Step Solution:

1. **Initial Drop**:
- The object is dropped from a height \( h \).
- The potential energy at height \( h \) is converted into kinetic energy just before it hits the ground.
- The potential energy (PE) at height \( h \) is given by:
\[ PE = mgh \]
- When it hits the ground, all this potential energy converts to kinetic energy (KE):
\[ KE = \frac{1}{2} mv^2 = mgh \]
- From this, we can find the velocity just before impact:
\[ v = \sqrt{2gh} \]

2. **First Impact**:
- Upon hitting the ground, the object loses 50% of its kinetic energy.
- Therefore, the kinetic energy after the first impact is:
\[ KE' = \frac{1}{2} KE = \frac{1}{2} (mgh) = \frac{1}{2} mgh \]

3. **Height Reached After First Bounce**:
- The remaining kinetic energy is converted back into potential energy as it rises.
- The potential energy at the new height \( h_1 \) is:
\[ mgh_1 = \frac{1}{2} mgh \]
- Solving for \( h_1 \):
\[ h_1 = \frac{1}{2} h \]

4. **Subsequent Bounces**:
- After reaching height \( h_1 \), the object again loses 50% of its kinetic energy when it hits the ground.
- The kinetic energy after the second impact becomes:
\[ KE'' = \frac{1}{2} KE' = \frac{1}{4} mgh \]
- The height reached after the second bounce \( h_2 \) is:
\[ mgh_2 = \frac{1}{4} mgh \implies h_2 = \frac{1}{4} h \]

5. **Continuing the Pattern**:
- This pattern continues, where the height after each bounce is halved:
- \( h_3 = \frac{1}{8} h \)
- \( h_4 = \frac{1}{16} h \)
- And so on...

6. **Total Distance Covered**:
- The total distance covered by the object can be calculated as:
\[ \text{Total Distance} = h + 2\left(h_1 + h_2 + h_3 + \ldots\right) \]
- The heights \( h_1, h_2, h_3, \ldots \) form a geometric series:
\[ h_1 + h_2 + h_3 + \ldots = \frac{h}{2} + \frac{h}{4} + \frac{h}{8} + \ldots \]
- The first term \( a = \frac{h}{2} \) and the common ratio \( r = \frac{1}{2} \).
- The sum 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**:
- Therefore, the total distance covered is:
\[ \text{Total Distance} = h + 2S = h + 2h = 3h \]

### Conclusion:
The total distance covered by the object as \( t \) tends to infinity is \( 3h \).