A ball is dropped from a height of `20 m` and rebounds with a velocity which is `3/4` of the velocity with which it hits the ground. What is the time interval between the first and second bunces `(g=10m//s^(2))`

Options

A. `3` sec
B. `4` sec
C. `5` sec
D. `6` sec

Explanation

To solve the problem, we need to calculate the time interval between the first and second bounces of a ball dropped from a height of 20 m, which rebounds with a velocity that is \( \frac{3}{4} \) of the velocity with which it hits the ground.

### Step-by-Step Solution:

1. **Calculate the velocity just before hitting the ground:**
The ball is dropped from a height \( h = 20 \, m \). We can use the equation of motion to find the velocity just before it hits the ground:
\[ v = \sqrt{2gh} \]
where \( g = 10 \, m/s^2 \) (acceleration due to gravity).

Substituting the values:
\[ v = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \, m/s \]

2. **Calculate the rebound velocity:**
The rebound velocity \( v' \) is \( \frac{3}{4} \) of the velocity just before hitting the ground:
\[ v' = \frac{3}{4} \times 20 = 15 \, m/s \]

3. **Calculate the time taken to fall to the ground:**
The time \( t_1 \) taken to fall from height \( h \) can be calculated using the formula:
\[ h = \frac{1}{2} g t_1^2 \]
Rearranging gives:
\[ t_1^2 = \frac{2h}{g} \implies t_1 = \sqrt{\frac{2h}{g}} \]
Substituting the values:
\[ t_1 = \sqrt{\frac{2 \times 20}{10}} = \sqrt{4} = 2 \, s \]

4. **Calculate the time taken to rise after the rebound:**
The time \( t_2 \) taken to reach the maximum height after rebounding can be calculated using:
\[ v' = g t_2 \implies t_2 = \frac{v'}{g} \]
Substituting the values:
\[ t_2 = \frac{15}{10} = 1.5 \, s \]

5. **Calculate the total time interval between the first and second bounces:**
The total time interval \( T \) between the first and second bounces is the sum of the time taken to fall and the time taken to rise:
\[ T = t_1 + t_2 = 2 + 1.5 = 3.5 \, s \]

### Final Answer:
The time interval between the first and second bounces is \( 3.5 \, s \).