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A tennis ball is dropped from a height of 9.8 m onto the floor. It rebounds to a height of 5.0 m. The ball is in contact with the floor for 0.2 s. Find the average acceleration during contact (in m/s²). [g = 10 m/s²]
- 120
- 240
- 80
- 60
Correct answer: 120
Solution
v_before = sqrt(2*10*9.8) = sqrt(196) = 14 m/s (downward). v_after = sqrt(2*10*5) = sqrt(100) = 10 m/s (upward). Taking upward as positive: delta_v = 10 - (-14) = 24 m/s. Average acceleration = 24/0.2 = 120 m/s².
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