A ball is dropped from a height h on to a floor . If the cofficient of restitution is e, calculate the height the ball first rebounds ?

Options

A. `e^2h`
B. `eh^2`
C. `e^4h`
D. `h/e^4`

Explanation

To solve the problem of finding the height to which a ball rebounds after being dropped from a height \( h \) with a coefficient of restitution \( e \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- The ball is dropped from a height \( h \).
- When the ball reaches the ground, it has a certain velocity due to gravitational acceleration.

2. **Calculate the Velocity Just Before Impact**:
- The velocity of the ball just before it hits the ground can be calculated using the equation of motion:
\[ v = \sqrt{2gh} \]
- Here, \( g \) is the acceleration due to gravity.

3. **Apply the Coefficient of Restitution**:
- The coefficient of restitution \( e \) relates the velocity of the ball just after the impact to the velocity just before the impact:
\[ v' = e \cdot v \]
- Substituting the value of \( v \):
\[ v' = e \cdot \sqrt{2gh} \]

4. **Calculate the Maximum Height of Rebound**:
- The maximum height \( h' \) to which the ball rebounds can be found using the kinetic energy at the moment just after the impact:
\[ h' = \frac{(v')^2}{2g} \]
- Substitute \( v' \) into the equation:
\[ h' = \frac{(e \cdot \sqrt{2gh})^2}{2g} \]
- Simplifying this:
\[ h' = \frac{e^2 \cdot 2gh}{2g} \]
- The \( 2g \) cancels out:
\[ h' = e^2 \cdot h \]

5. **Final Result**:
- Therefore, the height to which the ball first rebounds is:
\[ h' = e^2 \cdot h \]