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 ms -2 )

Question

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 ms -2 ) (A) 3.0 ms -1 (B) 3.50 ms -1 (C) 2.0 ms -1 (D) 2.50 ms -1

Tardigrade Admin · Accepted Answer

v 0=√2 gh v = e √2 gh =√2 gh ⇒ e =0.9 S = h +2 e 2 h +2 e 4 h + ldots ldots ldots t=√(2 h/g)+2 e √(2 h/g)+2 e2 √(2 h/g)+ ldots ldots . . v av =( S / t )=2.5 m / s

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

$3.0 m s^{- 1}$

$3.50 m s^{- 1}$

$2.0 m s^{- 1}$

$2.50 m s^{- 1}$

$v_{0} = 2 g h$
$v = e 2 g h = 2 g h$
$\Rightarrow e = 0.9$
$S = h + 2 e^{2} h + 2 e^{4} h + \dots\dots\dots$
$t = \frac{2 h}{g} + 2 e \frac{2 h}{g} + 2 e^{2} \frac{2 h}{g} + \dots\dots ..$
$v_{a v} = \frac{S}{t} = 2.5 m / s$

Q. A rubber ball is released from a height of 5m above the floor. It bounces back repeatedly, always rising to 10081​ of the height through which it falls. Find the average speed of the ball. (Take g=10ms−2 )

3.0ms−1

3.50ms−1

2.0ms−1

2.50ms−1