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```        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!!!```
7 years ago

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))
```
7 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.
```
7 years ago
Chetan Mandayam Nayakar
312 Points
```										both u and v are relative to ground
```
7 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.

piyush agrawal
```
7 years ago
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