A ball is dropped from a height h on the ground If the coefficient of restitution is e, the height to which the ball goes up after it rebounds for the `nth time is

Options

A. `he^(2n)`
B. `he^(n)`
C. `(e^(2n))/(h)`
D. `(h)/(e^(2n))`

Explanation

To solve the problem of determining 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. **Understanding the Coefficient of Restitution**:
The coefficient of restitution \( e \) is defined as the ratio of the velocity of separation to the velocity of approach. When the ball strikes the ground, it loses some of its velocity, and this loss is quantified by \( e \).

2. **Initial Velocity Calculation**:
When the ball is dropped from height \( h \), the velocity \( v_0 \) just before it hits the ground can be calculated using the equation of motion:
\[ v_0 = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity.

3. **Velocity After First Impact**:
After the first impact with the ground, the velocity \( v_1 \) with which the ball rebounds is given by:
\[ v_1 = e \cdot v_0 = e \cdot \sqrt{2gh} \]

4. **Height After First Rebound**:
The height \( h_1 \) to which the ball rebounds after the first impact can be calculated using the kinetic energy at the point of rebound:
\[ h_1 = \frac{v_1^2}{2g} = \frac{(e \cdot \sqrt{2gh})^2}{2g} = \frac{e^2 \cdot 2gh}{2g} = e^2 \cdot h \]

5. **Generalizing for nth Rebound**:
Following the same logic, after the second impact, the height \( h_2 \) to which the ball rebounds is:
\[ h_2 = e^2 \cdot h_1 = e^2 \cdot (e^2 \cdot h) = e^4 \cdot h \]
Continuing this pattern, after \( n \) rebounds, the height \( h_n \) can be expressed as:
\[ h_n = e^{2n} \cdot h \]

### Final Result:
Thus, the height to which the ball goes up after it rebounds for the \( n \)th time is:
\[ h_n = e^{2n} \cdot h \]