A heavy man , slips from a roof of a high building of height ′H′. When he was at a height of ′h′ he realised that the ground beneath him was pretty hard. To avoid injury, he thrown away his suitcase of mass ′m′. After doing this,he prevented his falling on the hard ground and fell on a pond. The mass of the man was ′M′. The distance of the pond from the hard ground was ′x′.And he just avoided from falling on the hard ground. Find the velocity required to throw away the suitcase as the man don′t fall on the hard ground.(There was no horizontal velocity of the suitcase−man system)


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Question Number 105037 by Dwaipayan Shikari last updated on 25/Jul/20

$A h e a v y m a n, s l i p s f r o m a r o o f o f a h i g h b u i l d i n g o f$ $h e i g h t^{'} H^{'} . W h e n h e w a s a t a h e i g h t o f^{'} h^{'} h e r e a l i s e d t h a t t h e$ $g r o u n d b e n e a t h h i m w a s p r e t t y h a r d . T o a v o i d i n j u r y, h e t h r o w n$ $a w a y h i s s u i t c a s e o f m a s s^{'} m^{'} . A f t e r d o i n g t h i s, h e p r e v e n t e d h i s$ $f a l l i n g o n t h e h a r d g r o u n d a n d f e l l o n a p o n d . T h e m a s s o f t h e$ $m a n w a s^{'} M^{'} . T h e d i s t a n c e o f t h e p o n d f r o m t h e h a r d g r o u n d w a s$ $^{'} x^{'} . A n d h e j u s t a v o i d e d f r o m f a l l i n g o n t h e h a r d g r o u n d .$ $F i n d t h e v e l o c i t y r e q u i r e d t o t h r o w a w a y t h e s u i t c a s e a s t h e m a n$ ${d o n}^{'} t f a l l o n t h e h a r d g r o u n d . (T h e r e w a s n o h o r i z o n t a l v e l o c i t y o f t h e$ $s u i t c a s e - m a n s y s t e m)$
	Answered by mr W last updated on 25/Jul/20

	$t i m e f o r t h e f a l l f r o m h e i g h t h t o$ $g r o u n d :$ $t = \sqrt{\frac{2 H}{g}} - \sqrt{\frac{2 (H - h)}{g}} = \sqrt{\frac{2 H}{g}} (1 - \sqrt{1 - \frac{h}{H}})$ $h o r i z o n t a l v e l o c i t y r e q u i r e d f o r t h e$ $m a n, i n o r d e r n o t t o f a l l o n t h e h a r d$ $g r o u n d :$ $u_{M} = \frac{x}{t} = \frac{x}{(1 - \sqrt{1 - \frac{h}{H}}) \sqrt{\frac{2 H}{g}}}$ $s a y t h e h o r i z o n t a l v e l o c i t y o f$ $s u i t c a s e i s u_{S},$ ${m u}_{S} = {M u}_{M}$ $\Rightarrow u_{S} = \frac{M}{m} u_{M}$ $r e q u i r e d v e l o c i t y t o t h r o w a w a y t h e$ $s u i t c a s e i s$ $v = u_{S} + u_{M} = (1 + \frac{M}{m}) u_{M}$ $\Rightarrow v = (1 + \frac{M}{m}) \frac{x}{(1 - \sqrt{1 - \frac{h}{H}}) \sqrt{\frac{2 H}{g}}}$
	Commented by Dwaipayan Shikari last updated on 25/Jul/20

	$G r e a t s i r!$
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