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

Correct answer: B. 1.5 s aligned Velocity on hitting the surface & = 2 × 9.8 × 4.9 & =9.8 m / s aligned Velocity after first bounce, v=(3)/(4) × 9.8Time taken from first bounce to the second bounce = (2 v)/(g)= 2 × (3)/(4) × 9.8 × (1)/(9.8)= 1.5 s

BITSAT 2024

Exam: BITSAT
Year: 2024
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

BITSAT 2024 · Physics · Center of Mass, Momentum and Collision

A ball falling freely from a height of 4.9 m / s , hits a...

A ball falling freely from a height of

4.9 \mathrm{ m} / \mathrm{s}

, hits a horizontal surface. If

e=\frac{3}{4}

, then the ball will hit the surface, second time after

Options

A1.0 s
B1.5 s
C2.0 s
D3.0 s

Correct answer

B. 1.5 s

Step-by-step solution

\begin{aligned} \text{Velocity on hitting the surface } & =\sqrt{2 \times 9.8 \times 4.9} \\ & =9.8 \mathrm{ m} / \mathrm{s} \end{aligned}

Velocity after first bounce,

v=\frac{3}{4} \times 9.8Time 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

\mathrm{ s}

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