A ball which is at rest is dropped from a height h metre. As it bounces off the floor its speed is 80% of what it was just before touching the ground. The ball will then rise to nearly a height :

Question

A ball which is at rest is dropped from a height h metre. As it bounces off the floor its speed is 80% of what it was just before touching the ground. The ball will then rise to nearly a height : (A) 0.94 h (B) 0.80 h (C) 0.75 h (D) 0.64 h

Tardigrade Admin · Accepted Answer

v=√2gh ...(1) After rebounce, v2=u2-2gh ⇒ u2=v2+2gh ∴ u2=2gh' ...(2) ∴ (v2/u2)=(2gh/2gh) ⇒ h'=h× (u2/v2)=h× ( (80/100) )2 =0.64 h

Q. A ball which is at rest is dropped from a height $h$ metre. As it bounces off the floor its speed is $80$ of what it was just before touching the ground. The ball will then rise to nearly a height :

0.94 h

0.80 h

0.75 h

0.64 h

0.50 h

$v = 2 g h$ ...(1)
After rebounce, $v^{2} = u^{2} - 2 g h$
$\Rightarrow$ $u^{2} = v^{2} + 2 g h$
$∴$ $u^{2} = 2 g h^{'}$ ...(2) $∴$
$\frac{v ^{2}}{u ^{2}} = \frac{2 g h}{2 g h}$
$\Rightarrow$ $h^{'} = h \times \frac{u ^{2}}{v ^{2}} = h \times (\frac{80}{100})^{2}$
$= 0.64 h$

Q. A ball which is at rest is dropped from a height h metre. As it bounces off the floor its speed is 80 of what it was just before touching the ground. The ball will then rise to nearly a height :