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))

Correct answer: (D) 2.50 m s^(-1). v_0=√2gh v=e√2gh=√2gh ⇒e=0.9 S=h+2e^2h+2e^4h+……… t=√(2h g)+2e√(2h g)+2e^2√(2h g)+……… v_av=s/t=2.5 m s^(-1)

Q61 / 90 · JEE Main 2021 (17 Mar Shift 2) · Physics

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

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

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

$\Rightarrow e = 0.9$

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

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

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

Hence, the correct answer is option: 'D'.