A ball falling freely from a height of 4.9 m / s , hits a horizontal surface. If e = 4 3 , then the ball will hit the surface, second time after

Velocity on hitting the surface = 2 × 9.8 × 4.9 = 9.8 m / s Velocity after first bounce, v = 4 3 × 9.8 Time taken from first bounce to the second bounce = g 2 v = 2 × 4 3 × 9.8 × 9.8 1 = 1.5 s

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PhysicsCenter of MassCollisionEasy2 minPYQ_2024

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A ball falling freely from a height of $4.9 m / s$ , hits a horizontal surface. If $e = \frac{3}{4}$ , then the ball will hit the surface, second time after

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$Velocity on hitting the surface = 2 \times 9.8 \times 4.9 = 9.8 m / s$ Velocity after first bounce, $v = \frac{3}{4} \times 9.8$ Time taken from first bounce to the second bounce $= \frac{2 v}{g} = 2 \times \frac{3}{4} \times 9.8 \times \frac{1}{9.8} = 1.5 s$

Stream:BITSATSubject:PhysicsTopic:Center of MassSubtopic:Collision

⏱ 2mℹ️ Source: PYQ_2024

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