To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If the ball is in contact with the floor for 12.0 ms, what is its average acceleration during that contact? Take `g= 98 m//s^2.`

Explanation

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

### Step 1: Identify the given values
- Initial height (h_initial) = 4.00 m
- Final height (h_final) = 2.00 m
- Time of contact (Δt) = 12.0 ms = 12.0 × 10^(-3) s
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Calculate the initial velocity just before impact
The initial velocity (V_initial) of the ball just before it hits the floor can be calculated using the formula for free fall:
\[ V_{initial} = \sqrt{2gh_{initial}} \]
Substituting the values:
\[ V_{initial} = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 4.00 \, \text{m}} \]
\[ V_{initial} = \sqrt{78.4} \]
\[ V_{initial} \approx 8.85 \, \text{m/s} \]

### Step 3: Calculate the final velocity just after rebound
The final velocity (V_final) of the ball just after it rebounds to a height of 2.00 m can be calculated similarly:
\[ V_{final} = \sqrt{2gh_{final}} \]
Substituting the values:
\[ V_{final} = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 2.00 \, \text{m}} \]
\[ V_{final} = \sqrt{39.2} \]
\[ V_{final} \approx 6.26 \, \text{m/s} \]

### Step 4: Determine the change in velocity
The change in velocity (ΔV) during the contact with the floor can be calculated as:
\[ \Delta V = V_{final} - (-V_{initial}) \]
(Note: The initial velocity is negative because it is directed downwards)
\[ \Delta V = 6.26 \, \text{m/s} - (-8.85 \, \text{m/s}) \]
\[ \Delta V = 6.26 + 8.85 \]
\[ \Delta V = 15.11 \, \text{m/s} \]

### Step 5: Calculate the average acceleration
The average acceleration (a_avg) can be calculated using the formula:
\[ a_{avg} = \frac{\Delta V}{\Delta t} \]
Substituting the values:
\[ a_{avg} = \frac{15.11 \, \text{m/s}}{12.0 \times 10^{-3} \, \text{s}} \]
\[ a_{avg} \approx 1259.17 \, \text{m/s}^2 \]

### Final Answer
The average acceleration of the tennis ball during its contact with the floor is approximately:
\[ a_{avg} \approx 1.26 \times 10^3 \, \text{m/s}^2 \]

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