A person of mass M holding a bag of mass m slips from the roof of a...

Let the total time taken to fall from the roof to the ground be $t$.

$t = \sqrt{\dfrac{2H}{g}}$

The time taken to fall from the roof to a height $h$ from the ground is $t_1$.

$t_1 = \sqrt{\dfrac{2(H-h)}{g}}$

The remaining time for the person to reach the ground is $t_2$.

$t_2 = t - t_1 = \dfrac{\sqrt{2H} - \sqrt{2(H-h)}}{\sqrt{g}}$

To land in the pond, the person must cover a horizontal distance $x$ in time $t_2$. The required horizontal velocity of the person $v_M$ is:

$v_M = \dfrac{x}{t_2} = \dfrac{x\sqrt{g}}{\sqrt{2H} - \sqrt{2(H-h)}}$

Initially, the horizontal momentum of the system (person + bag) is zero. By conservation of linear momentum in the horizontal direction:

$Mv_M = mv_m$

where $v_m$ is the horizontal velocity imparted to the bag in the ground frame.

$v_m = \dfrac{M}{m}v_M = \dfrac{Mx\sqrt{g}}{m\left[\sqrt{2H} - \sqrt{2(H-h)}\right]}$

Since no external force acts on the system in the horizontal direction, the center of mass of the system continues to fall along the initial vertical line of fall. Thus, the horizontal displacement of the center of mass is zero.

$Mx_M = mx_m$

Given that the person's horizontal displacement is $x_M = x$ (towards the pond), the bag's horizontal displacement $x_m$ is:

$x_m = \dfrac{M}{m}x$

Therefore, the bag will land at a distance $\dfrac{Mx}{m}$ to the left of the line of fall.