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!!!

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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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.

both u and v are relative to ground

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.

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piyush agrawal