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harsha 9128 Grade: 12
        

Q.in a gravity free space,a man of mass "M" standing at a height "h" above the ground throws a ball of mass "m" straight down with speed "u".when the ball reaches the ground the distance of the man from above the ground is?


ANS::


h(1+m/M)


HELP!!!

6 years ago

Answers : (4)

Chetan Mandayam Nayakar
312 Points
										

let speed of man be v. from principle of conservation of momentum, mu = Mv, v=mu/M, let time taken to reach ground be 't'. ut=h, t=h/u. therefore man moves vt=vh/u=mh/M, therefore, answer=h+(mh/M) =h(1+(m/M))


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6 years ago
harsha 9128
18 Points
										

thanks for the replies!!!


if the speed of man is"v" shouldnt the conservation of momentum be as follows


(M+m)v=mu....


or is it only applicable when the velocity of ball is "u" relative to man?


thanks again.

6 years ago
Chetan Mandayam Nayakar
312 Points
										

both u and v are relative to ground

6 years ago
PIYUSH AGRAWAL
31 Points
										

by conservation of momentum in vertical direction,


0=mu+Mv


or, v=-mu/M


so, velocity of man is mu/M in upward direction.


time taken by ball to reach ground,as acc. is zero, t=h/u


in this time, man moves up by a distance s=vt=mh/M


so, man is at height s+h=(1+m/M)h above the ground.


 


ask more....


piyush agrawal

6 years ago
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