Solve the following :

Solution:

When he throws a bag to left, momentum is conserved. $m v_{b}=M V_{M}$

$V_{b}=$ velocity of bag

$V_{M}=$ speed of man

Let pond start from $\left({ }^{x_{i}}\right)$

Time to reach height $\mathrm{h}=\sqrt{2(H-h) / g}=t_{h}$

Time to reach ground, $t_{g}=\sqrt{2 H / g}=t_{g}$

Time taken by man $=t_{g}-t_{h}=t_{M}$

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

COM of the system will be at $(0,0)$ after he reaches ground (let bag reach ${ }^{x_{1}}$ from origin) $\left({ }^{x_{1}}, 0\right.$ )

$\therefore 0=\frac{M \times x+m \times x_{1}}{M+m}$

$x_{1}=\frac{-M}{m} x$

From law of kinematics

$x_{M}-0=V_{M} t \quad\left\{x_{M}=x\right\}$

$x=V_{M} t_{M}=v\left(t_{g}-t_{h}\right)$

$V_{M}=\frac{x}{t_{g}-t_{h}}=x \sqrt{\frac{g}{2}}(\sqrt{H}-\sqrt{H-h})^{-1}$

$v_{b}=\frac{M}{m} V_{M}=\frac{M x}{m} \sqrt{\frac{g}{2}} \frac{1}{\sqrt{H}-\sqrt{H-h}}$