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) 2.50 ms ⁻¹ (b) 3.0 ms ⁻¹ (c) 2.0 ms ⁻¹ (d) 3.50 ms ⁻¹

Answer: (a). Height ratio e² = 0.81 ⇒ speed ratio e = 0.9 per bounce. Total distance = h(1 + e²)/(1 - e²) = 5 × (1.81)/(0.19) ≈ 47.6 m First fall time t₀ = √(2h/g) = 1 s; total time = t₀(1 + e)/(1 - e) = (1.9)/(0.1) = 19 s Average speed = (47.6)/(19) ≈ 2.5 m/s So the answer is (a) 2.50 m/s.

A rubber ball is released from a height of 5 m above the floor. It… | JEE Main 2021 (17 Mar, Shift 2)

Q38JEE Main 2021 · 17 MAR · SHIFT 2Difficult

Motion Under Gravity

A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to $\frac{81}{100}$ of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms ${}^{-2}$ )

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Answer: (a)

Height ratio $e^2 = 0.81 \Rightarrow$ speed ratio $e = 0.9$ per bounce.
Total distance $= h\frac{1 + e^2}{1 - e^2} = 5 \times \frac{1.81}{0.19} \approx 47.6\text{ m}$
First fall time $t_0 = \sqrt{\frac{2h}{g}} = 1\text{ s}$ ; total time $= t_0\frac{1 + e}{1 - e} = \frac{1.9}{0.1} = 19\text{ s}$
Average speed $= \frac{47.6}{19} \approx 2.5\text{ m/s}$
So the answer is (a) 2.50 m/s.

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