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 of finding the total time after which a ball dropped from height 'H' comes to rest, given the coefficient of restitution 'e', we can follow these steps:

### Step 1: Determine the initial velocity of the ball just before impact
When the ball is dropped from height 'H', it falls under the influence of gravity. The velocity \( V \) just before it hits the ground can be calculated using the equation of motion:

\[ V = \sqrt{2gH} \]

where \( g \) is the acceleration due to gravity.

**Hint:** Use the formula for velocity derived from the conservation of energy or the equations of motion for free fall.

### Step 2: Calculate the time taken to fall to the ground
The time \( T \) taken to fall from height \( H \) to the ground can be found using the equation:

\[ T = \frac{V}{g} = \frac{\sqrt{2gH}}{g} = \sqrt{\frac{2H}{g}} \]

**Hint:** Remember that the initial velocity \( u = 0 \) when the ball is dropped.

### Step 3: Determine the height to which the ball rebounds
After the first impact, the ball rebounds to a height \( H_1 \) given by:

\[ H_1 = e^2 H \]

where \( e \) is the coefficient of restitution. The velocity with which it rebounds is \( V_1 = eV = e\sqrt{2gH} \).

**Hint:** The coefficient of restitution relates the velocities before and after the collision.

### Step 4: Calculate the time taken to reach the height \( H_1 \)
The time \( T_1 \) taken to reach the height \( H_1 \) can be calculated similarly:

\[ T_1 = \frac{V_1}{g} = \frac{e\sqrt{2gH}}{g} = e\sqrt{\frac{2H}{g}} \]

**Hint:** Use the same approach as in Step 2, but now with the rebound velocity.

### Step 5: Repeat for subsequent bounces
The ball will continue to bounce, each time reaching a height reduced by a factor of \( e^2 \). The time for each subsequent bounce can be expressed as:

- For the second rebound: \( T_2 = e^2 \sqrt{\frac{2H}{g}} \)
- For the third rebound: \( T_3 = e^3 \sqrt{\frac{2H}{g}} \)
- And so on...

**Hint:** Observe the pattern in the time taken for each bounce.

### Step 6: Sum the total time taken until the ball comes to rest
The total time \( T_{total} \) is the sum of the time taken for the fall and all the rebounds:

\[ T_{total} = T + T_1 + T_2 + T_3 + \ldots \]

This can be expressed as:

\[ T_{total} = \sqrt{\frac{2H}{g}} + e\sqrt{\frac{2H}{g}} + e^2\sqrt{\frac{2H}{g}} + e^3\sqrt{\frac{2H}{g}} + \ldots \]

This is a geometric series with the first term \( a = \sqrt{\frac{2H}{g}} \) and common ratio \( r = e \).

**Hint:** Use the formula for the sum of an infinite geometric series \( S = \frac{a}{1 - r} \) where \( |r| < 1 \).

### Step 7: Calculate the total time using the geometric series formula
The total time can be calculated as:

\[ T_{total} = \sqrt{\frac{2H}{g}} \left( \frac{1}{1 - e} \right) \]

Thus, the final expression for the total time after which the ball comes to rest is:

\[ T_{total} = \sqrt{\frac{2H}{g}} \cdot \frac{1 + e}{1 - e} \]

### Final Answer
The total time after which the ball comes to rest is:

\[ T_{total} = \sqrt{\frac{2H}{g}} \cdot \frac{1 + e}{1 - e} \]