A ball is dropped from height 5m. The time after which ball stops rebounding if coefficient of restitution between ball and ground `e=1//2`, is

Answer

To solve the problem step by step, we will follow the outlined process:

### Step 1: Understand the Problem
We need to find the total time after which a ball, dropped from a height of 5 meters, stops rebounding when the coefficient of restitution (e) between the ball and the ground is 1/2.

### Step 2: Calculate the Time of Flight (t)
The time of flight for a ball dropped from a height (h) can be calculated using the formula:
\[ t = \sqrt{\frac{2h}{g}} \]
where \( g \) is the acceleration due to gravity, approximately \( 10 \, \text{m/s}^2 \).

Given:
- \( h = 5 \, \text{m} \)
- \( g = 10 \, \text{m/s}^2 \)

Substituting the values:
\[ t = \sqrt{\frac{2 \times 5}{10}} = \sqrt{1} = 1 \, \text{s} \]

### Step 3: Calculate the Total Time Until the Ball Stops Rebounding (t')
To find the total time until the ball stops rebounding, we use the formula:
\[ t' = t \cdot \frac{1 + e}{1 - e} \]
where \( e \) is the coefficient of restitution.

Given:
- \( e = \frac{1}{2} \)
- \( t = 1 \, \text{s} \)

Substituting the values:
\[ t' = 1 \cdot \frac{1 + \frac{1}{2}}{1 - \frac{1}{2}} = 1 \cdot \frac{\frac{3}{2}}{\frac{1}{2}} = 1 \cdot 3 = 3 \, \text{s} \]

### Conclusion
The ball stops rebounding after a total time of \( 3 \, \text{s} \).

### Final Answer
The ball stops rebounding after **3 seconds**.

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