A ball is dropped from a height of 22.5 meter on a fixed horizontal plane. If e = 2 / 5, then it will stop rebounding after

Options

A. 5 sec
B. 6 sec
C. 7 sec
D. 8 sec

Answer

To solve the problem of how many rebounds a ball will make when dropped from a height of 22.5 meters with a coefficient of restitution \( e = \frac{2}{5} \), we can follow these steps:

### Step 1: Determine the initial drop height and the gravitational acceleration
We know the initial height \( h = 22.5 \) meters and the acceleration due to gravity \( g \approx 9.81 \, \text{m/s}^2 \).

### Step 2: Calculate the time taken to fall to the ground
Using the equation of motion for free fall:
\[ h = \frac{1}{2} g t_0^2 \]
Rearranging gives:
\[ t_0 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 22.5}{9.81}} \approx \sqrt{4.58} \approx 2.14 \, \text{s} \]

### Step 3: Calculate the velocity just before impact
Using the equation:
\[ v = gt_0 = 9.81 \times 2.14 \approx 21.0 \, \text{m/s} \]

### Step 4: Calculate the velocity after the first rebound
The velocity after the rebound can be calculated using the coefficient of restitution:
\[ v' = e \cdot v = \frac{2}{5} \cdot 21.0 \approx 8.4 \, \text{m/s} \]

### Step 5: Calculate the height after the first rebound
Using the formula for height based on the rebound velocity:
\[ h' = \frac{v'^2}{2g} = \frac{(8.4)^2}{2 \cdot 9.81} \approx \frac{70.56}{19.62} \approx 3.6 \, \text{m} \]

### Step 6: Repeat the process for subsequent rebounds
We can repeat the process for each subsequent rebound, calculating the new height after each rebound using the same method. The height after each rebound will be:
\[ h_n = e^2 \cdot h_{n-1} \]
Where \( h_0 = 22.5 \, \text{m} \) and \( e = \frac{2}{5} \).

### Step 7: Determine when the ball stops rebounding
The ball will stop rebounding when the height becomes negligible. We can find the number of rebounds by calculating how many times the height can be multiplied by \( e^2 \) before it becomes less than a small threshold (e.g., 0.01 m).

### Calculation of rebounds
1. First height: \( h_0 = 22.5 \)
2. First rebound height: \( h_1 = e^2 \cdot h_0 = \left(\frac{2}{5}\right)^2 \cdot 22.5 = \frac{4}{25} \cdot 22.5 = 3.6 \)
3. Second rebound height: \( h_2 = e^2 \cdot h_1 = \left(\frac{2}{5}\right)^2 \cdot 3.6 = \frac{4}{25} \cdot 3.6 = 0.576 \)
4. Third rebound height: \( h_3 = e^2 \cdot h_2 = \left(\frac{2}{5}\right)^2 \cdot 0.576 = \frac{4}{25} \cdot 0.576 = 0.09216 \)
5. Fourth rebound height: \( h_4 = e^2 \cdot h_3 = \left(\frac{2}{5}\right)^2 \cdot 0.09216 = \frac{4}{25} \cdot 0.09216 \approx 0.0147 \)

After the fourth rebound, the height is approximately 0.0147 m, which is still above our threshold. Continuing this process, we find that the ball will stop rebounding after a total of 5 rebounds.

### Final Answer
The ball will stop rebounding after **5 rebounds**.