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 (figure 9-E8). 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?
Total time man will take to reach the height h = t = time taken in falling through height H – time taken in falling through height (H-h)
Horizontal distance covered in this time = x
Horizontal velocity of the man, v2 = x/t
Conserving momentum in the horizontal direction, mv1 = Mv2
⇒ v1 = Mv2/m
Where v1 = the horizontal velocity with which the bag is thrown towards left
∴Distance the bag travels = v1t = Mx/m
And, the minimum horizontal velocity imparted to the bag
