A ball of mass M falls from a height on a floor for which the coefficient of restitution is e. The height attained by the ball after two rebounds is

Options

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

Explanation

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

### Step 1: Determine the velocity just before the first impact
When the ball falls from height \( h \), we can use the equation of motion to find its velocity just before it hits the ground. The equation is given by:

\[ V^2 = U^2 + 2gh \]

Where:
- \( V \) is the final velocity just before impact,
- \( U \) is the initial velocity (which is 0 when dropped),
- \( g \) is the acceleration due to gravity,
- \( h \) is the height from which it is dropped.

Since the ball is dropped, \( U = 0 \):

\[ V^2 = 0 + 2gh \implies V = \sqrt{2gh} \]

### Step 2: Calculate the velocity after the first rebound
Using the coefficient of restitution \( e \), the velocity of the ball after the first rebound will be:

\[ V' = e \cdot V = e \cdot \sqrt{2gh} \]

### Step 3: Determine the height attained after the first rebound
The height attained after the first rebound can be calculated using the same equation of motion. The ball will rise to a height \( h_1 \) given by:

\[ V'^2 = 2gh_1 \]

Setting \( V' = e \cdot \sqrt{2gh} \):

\[ (e \cdot \sqrt{2gh})^2 = 2gh_1 \]

This simplifies to:

\[ e^2 \cdot 2gh = 2gh_1 \implies h_1 = e^2 h \]

### Step 4: Calculate the velocity after the second rebound
After reaching the height \( h_1 \), the ball will fall again and hit the ground with the same velocity it had just before the first rebound. Therefore, just before the second impact, the velocity will be:

\[ V'' = \sqrt{2gh_1} = \sqrt{2g(e^2h)} = e \cdot \sqrt{2gh} \]

Using the coefficient of restitution again, the velocity after the second rebound will be:

\[ V'' = e \cdot V' = e^2 \cdot \sqrt{2gh} \]

### Step 5: Determine the height attained after the second rebound
Now, we calculate the height attained after the second rebound \( h_2 \):

\[ (V'')^2 = 2gh_2 \]

Substituting \( V'' = e^2 \cdot \sqrt{2gh} \):

\[ (e^2 \cdot \sqrt{2gh})^2 = 2gh_2 \]

This simplifies to:

\[ e^4 \cdot 2gh = 2gh_2 \implies h_2 = e^4 h \]

### Final Result
The total height attained by the ball after two rebounds is:

\[ h_2 = e^4 h \]