A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to 81/100 of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms⁻²)
A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to 81/100 of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms⁻²)
Option 1 -
2.0 ms⁻¹
Option 2 -
2.50 ms⁻¹
Option 3 -
3.0 ms⁻¹
Option 4 -
3.50 ms⁻¹
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1 Answer
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Correct Option - 2
Detailed Solution:For a bouncing object with initial height h = 5m and coefficient of restitution e = 0.9 (so e² =0.81):
Total distance traveled, d = h + 2e²h + 2e? h + . = h * (1 + e²) / (1 - e²)
Total time taken, t = √* (2h/g)* + 2√* (2e²h/g)* + 2√* (2e? h/g)* + . = √* (2h/g)* * (1 + e) / (1 - e)
Average speed = d/t = √* (gh/2)* * (1 + e²) / (1 + e)² = 5 * (1.81) / (1.9)² = 2.50 m/s
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