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 then rise to nearly a height

Question

Dr. Abubacker · Accepted Answer

(d) $v = \sqrt{2\ gh}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
After rebounce, $v^{2} = u^{2} - 2gh$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u^{2} = v^{2} + 2\ gh'$
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u^{2} = 2\ gh^{'}\ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
$	herefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{v^{2}}{u^{2}} = \frac{2gh}{2gh^{'}}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h^{'} = h 	imes \frac{u^{2}}{v^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = h 	imes \left( \frac{80}{100} ight)^{2} = 0.64\ h$

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 then rise to nearly a height

Explanation

Reviewed by Dr. Abubacker

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