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To test the quality of a tennis ball, yo...

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.`

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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: \[ a_{avg} \approx 1.26 \times 10^3 \, \text{m/s}^2 \] ---

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 ...
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