 # 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!!!

12 years ago

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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12 years ago

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.

12 years ago

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

3 years ago
Here there is no external unbalanced force, so the law of conservation of momentum can be applied.
Pi=Pf
Initially, man at rest.
Pi= 0
When he throws the ball, the ball will hit the ground but as there is no gravity, the man will fly upwards.
Pf=Mv+(m (-u))
Where M is the mass of man, v is the velocity of man
m is the mass of ball, u is the velocity of ball
0= Mv+(-mu)
Mv=my
V= mu÷M      ( 1)
Consider after the ball is thrown, the man is flown up by distance x
As velocity is distance by time
V= x/t             (2)
From 1 and 2
mu/M=x/t
x= (mu/M) ×t
For ball,
u= h/t
X= mu/t  ×  h/u
So man will be at a distance of h+ x
h+x= h+ mh/M
= h (1+m/M)