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 m s - 2 )

v 0 = 2 g h v = e 2 g h = 2 g h ⇒ e = 0 . 9 S = h + 2 e 2 h + 2 e 4 h + … … … t = 2 h g + 2 e 2 h g + 2 e 2 2 h g + … … … v av = s t = 2 . 5 m s - 1

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PhysicsCenter of MassCollisionHard2 minPYQ_2021

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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 $g = 10 {m s}^{- 2}$ )

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Answer:

Solution:

$v_{0} = \sqrt{2 g h}$

$v = e \sqrt{2 g h} = \sqrt{2 g h}$

$\Rightarrow e = 0.9$

$S = h + 2 e^{2} h + 2 e^{4} h + \dots \dots \dots$

$t = \sqrt{\frac{2 h}{g}} + 2 e \sqrt{\frac{2 h}{g}} + 2 e^{2} \sqrt{\frac{2 h}{g}} + \dots \dots \dots$

$v_{av} = \frac{s}{t} = 2.5 m s^{- 1}$

Stream:JEESubject:PhysicsTopic:Center of MassSubtopic:Collision

⏱ 2mℹ️ Source: PYQ_2021

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