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 - <p>2.0 ms⁻¹</p>
Option 2 - <p>2.50 ms⁻¹</p>
Option 3 - <p>3.0 ms⁻¹</p>
Option 4 - <p>3.50 ms⁻¹</p>
27 Views|Posted 5 months ago
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A
5 months ago
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 =

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Physics Work, Energy and Power 2021

Physics Work, Energy and Power 2021

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