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 below 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.If the man just succeeds to avoid the hard ground,where will the bag land?

Answer 1

Mass of man=M

Initial velocity=0

Mass of bag=m

Let the man throws the bag towards left with a velocity v.

So,there is no external force in the horizontal direction.

The momentum will be conserved.Let he goes right with a velocity mv=MV

$\Rightarrow V=\left(\frac{mv}{M}\right)$

$\Rightarrow v=\left(\frac{MV}{m}\right)...\left(i\right)$

Let the total time he will take to reach ground

$=\sqrt{\frac{2H}{g}}=t_1$

Let the total time he will take to reach the height h.

$=t_2=\sqrt{\frac{2(H-h)}{g}}$

Then the time of his flying $=t_1-t_2$

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

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

Within this time he reaches the ground in the pond covering a horizontal distance

$\Rightarrow x=V\times t$

$\Rightarrow V=\frac{x}{t}$

$v=\frac{M}{m}\frac{x}{t}$

$=\frac{M}{m}\times \frac{\sqrt{g}}{\sqrt{2}(\sqrt{H}-\sqrt{H-h})}x$

As there is no external force in horizontal direction,the x-coordinate of CM will remian at that position.

$\Rightarrow 0=\frac{M\times \left(x\right)+m\times x_1}{M+m}$

$\Rightarrow x_1=\frac{M}{m}x$

$\therefore$ The bag will reach the bottom at a distance $\left(\frac{M}{m}\right)x$ towards left of the line it falls.