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 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?




6 years ago


Answers : (4)


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

thanks for the replies!!!

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


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

thanks again.

6 years ago

both u and v are relative to ground

6 years ago

by conservation of momentum in vertical direction,


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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