Question

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 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?

Answer 1

Since the falling man with the bag has zero momentum in the horizontal direction when he throws the bag horizontally the momenta of man and the bag will be equal and opposite. The first thing to know here is the time taken by the man (t) to reach the ground in falling through height h.
t = time in falling through H - time in falling through (H-h)
=√(2H/g)-√{2(H-h)/g} 
In this time the man covers a horizontal distance of x, so the horizontal velocity of the man
V=x/t = x/[√(2H/g)-√{2(H-h)/g}]
If the horizontal velocity of the bag is v,
then, mv = MV
v = MV/m
=Mx/[√(2H/g)-√{2(H-h)/g}].m
=Mx√g/m[√(2H)-√{2(H-h)}] 
The distance traveled by the bag with this velocity in time t
=vt
=Mx√g/m[√(2H)-√{2(H-h)}]*[√(2H/g)-√{2(H-h)/g}]
=Mx/m away from the line of fall opposite to the direction of the pond.

Answer 2

Thanks for asking the question!

ANSWER::

See figure for better understanding.

Mass of man = M     Initial velocity = 0

Mass of bag = m

Let the man throw the bag towards left with a velocity v towards left.
So , we can say that there is no external force in the horizontal direction.
Momentum will be conserved.

Let man go towards right with velocity 

mv = MV 

V = mv / M

v = MV / m ................... Equation - 1

Let the total time man will take to reach the ground = √(2H/g) = t₁

Let the total time man will take to reach the height h = √[2(H-h)/g] = t₂

Time of flight of the man = t₁-t₂ = √(2H/g) - √[2(H-h)/g] = √(2/g)[√H - √(H-h)]

In this time he reaches the ground in the pond covering a horizontal distance x .

x = V x t

V = x/t

v = Mx / mt = Mx√g / m[√2(√H - √(H-h)]

There is no external force in horizontal direction so , the x-coordinate of centre of mass will remain at that position.

0 = (M x (x) + m x x₁) / (M+m)

x₁ = - Mx/m

Therefore , the bag will reach bottom at a distance (Mx/m) towards left of the line it falls.

Hope it helps!