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1.A man of mass M having a bag of mass m slips from the roof of a tall building of height H and starts falling vertically. When at a height h from the ground he notices that the ground below him is pretty hard, but there is a pond at a horizontal distance x from the line of fall. In order to save himself he throws the bag horizontally (with respect to himself) in the direction opposite to the pond. Calculate the minimum horizontal velocity imparted to the bag so that the man lands in the water. If the man just succeeds to avoid the hard ground, where will the bag land?
let u,v be horizontal velocities of man and bag
balancing moments gives : Mu = mv
and h=1/2.g.t2 , x = ut
from all 3 eqns , we have hz. velocity of bag is v = Mx /[m.(2h/g)1/2 ]
now tha bag will land at , y = vt
on solving all eqns we have
bag lands at : y = Mx/m
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regards
Ramesh
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