A ball is dropped onto the floor from a height of 20 m. It rebounds to a height of 10 m. If the ball is in contact with the floor for 0.1 seconds, what is the average acceleration during contact?

Options

A. `142 ms^(-2)`
B. `285 ms^(-2)`
C. `338 ms^(-2)`
D. `564 ms^(-2)`

Explanation

To solve the problem of finding the average acceleration of a ball during its contact with the floor, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Heights**:
- The ball is dropped from a height \( H_1 = 20 \, \text{m} \).
- It rebounds to a height \( H_2 = 10 \, \text{m} \).

2. **Determine the Initial and Final Velocities**:
- When the ball is dropped from height \( H_1 \), its initial velocity \( V_1 \) just before hitting the ground can be calculated using the formula:
\[ V_1 = \sqrt{2gH_1} \]
where \( g \) is the acceleration due to gravity (\( g \approx 9.8 \, \text{m/s}^2 \)).
- Substitute \( H_1 \):
\[ V_1 = \sqrt{2 \times 9.8 \times 20} = \sqrt{392} \approx 19.8 \, \text{m/s} \]

- When the ball rebounds to height \( H_2 \), its final velocity \( V_2 \) just after leaving the ground can be calculated similarly:
\[ V_2 = \sqrt{2gH_2} \]
Substitute \( H_2 \):
\[ V_2 = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} \approx 14.0 \, \text{m/s} \]

3. **Set the Sign Convention**:
- We take upward direction as positive and downward direction as negative.
- Therefore, \( V_1 \) is negative and \( V_2 \) is positive:
\[ V_1 = -19.8 \, \text{m/s}, \quad V_2 = 14.0 \, \text{m/s} \]

4. **Calculate the Average Acceleration**:
- The average acceleration \( a \) during the contact time \( t \) (which is given as \( 0.1 \, \text{s} \)) can be calculated using the formula:
\[ a = \frac{V_f - V_i}{t} \]
- Here, \( V_f = V_2 \) and \( V_i = V_1 \):
\[ a = \frac{14.0 - (-19.8)}{0.1} \]
\[ a = \frac{14.0 + 19.8}{0.1} = \frac{33.8}{0.1} = 338 \, \text{m/s}^2 \]

5. **Final Result**:
- The average acceleration during the contact with the floor is approximately \( 338 \, \text{m/s}^2 \).