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). 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.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Let the horizontal velocity imparted to the bag with respect to ground be $u_b$ (towards left) and Let the velocity of man be $u_m$ (towards right). As there is no net External force in horizontal direction, its linear momentum is covered in this direction.

$-m{V}_b+M{V}_m=0$

$v_b=\frac{M}{m}{V}_m$ ....(1) Towards left as assumed

Now let us final the minimum horizontal velocity required at P (h height above the ground) in order to cover k units.

As there is no force along this direction, $u_M$ will not change. IF 't' is the time for man and bag to reach the

ground from point p ( as shown in figure ) and velocity of ( Man + Bag) at P

Assume downwards to be positive y axis

$V^2=O^2+2\times g\times (H-h)$

v = $\sqrt{2g(H-h)}$ ........(2)

Now,

S = ht + $\frac{1}{2}a{t}^2$

h = $\sqrt{2g(H-h)t}+\frac{1}{2}g{t}^2$

$\frac{1}{2}g{t}^2+\sqrt{2g(H-h)t}-h=0$

Solving for t,

$t=\frac{-\sqrt{2g(H-h)}\pm \sqrt{2g(H-h)+2gh}}{g}$

$=\frac{\sqrt{2g(H-h)}\pm \sqrt{2gH}}{g}$

But time cannot be -ve,

So, t = $\frac{\sqrt{2gH}-\sqrt{2g(H-h)}}{g}$

Using speed = $\frac{distance}{time}$

$\Rightarrow V_M=\frac{xg}{\sqrt{2gH}-\sqrt{2(H-h)}}$

put in (1)

$V_b=\frac{Mxg}{m\sqrt{g}\{\sqrt{2H}-\sqrt{2(H-h)}}$

or $V_b=\frac{Mx\sqrt{g}}{m\{\sqrt{2H}-\sqrt{2(H-h)}\}}$