A ball is dropped from height 'H' onto a horizontal surface. If the coefficient of restitution is 'e' then the total time after which it comes to rest is

Options

A. `sqrt((2H)/(g))((1-e)/(1+e))`
B. `sqrt((2H)/(g))((1+e)/(1-e))`
C. `sqrt((2H)/(g))((1+e^(2))/(1-e^(2)))`
D. `sqrt((2H)/(g))((1-e^(2))/(1+e^(2)))`

Answer

To solve the problem, we need to find the total time after which a ball dropped from a height \( H \) comes to rest after bouncing on a horizontal surface with a coefficient of restitution \( e \).

### 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. For a ball dropped from height \( H \), the height to which it rebounds after the first bounce can be expressed as:
\[ h_1 = e^2 H \]

2. **Time of Fall from Height \( H \):**
The time taken to fall from height \( H \) can be calculated using the formula for free fall:
\[ t_1 = \sqrt{\frac{2H}{g}} \]
where \( g \) is the acceleration due to gravity.

3. **Time of Rise to Height \( h_1 \):**
The time taken to rise to the height \( h_1 \) is:
\[ t_2 = \sqrt{\frac{2h_1}{g}} = \sqrt{\frac{2(e^2 H)}{g}} = e \sqrt{\frac{2H}{g}} \]

4. **Subsequent Bounces:**
After the first bounce, the ball will continue to bounce back to heights that are reduced by a factor of \( e^2 \) each time. The heights for the subsequent bounces can be expressed as:
\[ h_n = e^{2n} H \]
The time taken for each subsequent fall and rise can be calculated similarly:
\[ t_{n, fall} = \sqrt{\frac{2h_n}{g}} = e^n \sqrt{\frac{2H}{g}} \quad \text{and} \quad t_{n, rise} = e^n \sqrt{\frac{2H}{g}} \]

5. **Total Time Calculation:**
The total time \( T \) for the ball to come to rest is the sum of the time for all falls and rises:
\[ T = t_1 + t_2 + t_3 + \ldots \]
This can be expressed as:
\[ T = \sqrt{\frac{2H}{g}} + e \sqrt{\frac{2H}{g}} + e^2 \sqrt{\frac{2H}{g}} + \ldots \]
This is a geometric series with the first term \( a = \sqrt{\frac{2H}{g}} \) and common ratio \( r = e \).

6. **Sum of the Infinite Series:**
The sum of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
Thus, the total time \( T \) becomes:
\[ T = \frac{\sqrt{\frac{2H}{g}}}{1 - e} \]

### Final Result:
The total time after which the ball comes to rest is:
\[ T = \frac{\sqrt{2H}}{g(1 - e)} \]