A rubber ball is released from a height of 5m 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 )

Correct answer: D. 2 . 50 ms ^ - 1 v _ 0 = 2gh v = e 2gh = 2gh ⇒ e = 0 . 9 S = h + 2 e ^ 2 h + 2 e ^ 4 h + t = (2h)/(g) + 2e (2h)/(g) + 2 e ^ 2 (2h)/(g) + v _ av = (s)/(t) = 2 . 5m s ^ - 1

JEE Main 202117 MarchEvening Shift

Exam: JEE Main
Year: 2021
Date: 17 March
Shift: Evening Shift
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

JEE Main 2021 · 17 March · Evening Shift · Physics · Center of Mass, Momentum and Collision

A rubber ball is released from a height of 5m above the floor. It...

A rubber ball is released from a height of

5m

above the floor. It bounces back repeatedly, always rising to

\dfrac{81}{100}

of the height through which it falls. Find the average speed of the ball. (Take

g = 10{ms}^{- 2}

)

Options

A $3 . 0{ms}^{- 1}$
B $3 . 5{ms}^{- 1}$
C $2 . 0{ms}^{- 1}$
D $2 . 50{ms}^{- 1}$

Correct answer

D. $2 . 50{ms}^{- 1}$

Step-by-step solution

${v}_{0} = \sqrt{2gh}$

$v = e\sqrt{2gh} = \sqrt{2gh}$

$⇒ e = 0 . 9$

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

$t = \sqrt{\dfrac{2h}{g}} + 2e\sqrt{\dfrac{2h}{g}} + 2{e}^{2}\sqrt{\dfrac{2h}{g}} + \ldots \ldots \ldots$

${v}_{av} = \dfrac{s}{t} = 2 . 5m{s}^{- 1}$

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