A ball is dropped from a height h on a floor. The coefficient of restitution for the collision between the ball and the floor is e. The total distance covered by the ball before it comes to the rest.

Options

A. `h(1-2e^(2))`
B. `h[(1+e^(2))/(1-e^(2))]`
C. `h[(1-e^(2))/(1+e^(2))]`
D. `he^(2)`

Answer

To solve the problem of finding the total distance covered by a ball dropped from a height \( h \) before it comes to rest, given the coefficient of restitution \( e \), we can break down the solution into several steps.

### Step-by-Step Solution:

1. **Initial Drop**:
The ball is dropped from a height \( h \). When it hits the ground, it will have a velocity \( v \) just before the impact. Using the equation of motion:
\[ v^2 = u^2 + 2gh \]
where \( u = 0 \) (initial velocity), we find:
\[ v^2 = 2gh \quad \Rightarrow \quad v = \sqrt{2gh} \]

2. **Velocity After Collision**:
After the collision, the velocity of the ball just after it rebounds is given by the coefficient of restitution:
\[ u = e \cdot v = e \cdot \sqrt{2gh} \]

3. **Height After First Bounce**:
The ball will rise to a height \( h' \) after the first bounce. Using the equation of motion again:
\[ 0 = u^2 - 2gh' \quad \Rightarrow \quad h' = \frac{u^2}{2g} = \frac{(e \cdot \sqrt{2gh})^2}{2g} = \frac{e^2 \cdot 2gh}{2g} = e^2h \]

4. **Subsequent Bounces**:
After the first bounce, the ball will continue to bounce and rise to heights that are reduced by the factor of \( e^2 \) for each subsequent bounce. The heights after each bounce can be expressed as:
- First bounce height: \( h' = e^2h \)
- Second bounce height: \( h'' = e^4h \)
- Third bounce height: \( h''' = e^6h \)
- And so on...

5. **Total Distance Calculation**:
The total distance \( D \) covered by the ball includes the initial drop and the distances covered during the upward and downward motions for each bounce:
\[ D = h + 2(h' + h'' + h''') + \ldots \]
This can be expressed as:
\[ D = h + 2\left( e^2h + e^4h + e^6h + \ldots \right) \]

6. **Sum of Infinite Geometric Series**:
The series \( e^2h + e^4h + e^6h + \ldots \) is a geometric series with the first term \( a = e^2h \) and common ratio \( r = e^2 \):
\[ \text{Sum} = \frac{a}{1 - r} = \frac{e^2h}{1 - e^2} \]
Therefore, the total distance becomes:
\[ D = h + 2 \cdot \frac{e^2h}{1 - e^2} = h + \frac{2e^2h}{1 - e^2} \]

7. **Final Expression**:
Combining the terms, we can express the total distance covered by the ball before it comes to rest as:
\[ D = h \left( 1 + \frac{2e^2}{1 - e^2} \right) = h \left( \frac{1 - e^2 + 2e^2}{1 - e^2} \right) = h \left( \frac{1 + e^2}{1 - e^2} \right) \]

### Final Answer:
\[ D = h \left( \frac{1 + e^2}{1 - e^2} \right) \]