A small body of mass m is placed over a larger mass M whose surface...

Let $V$ be the common horizontal velocity of the small body $m$ and the larger mass $M$ when the small body breaks off. Since the surface becomes vertical at the top, the horizontal velocity of $m$ relative to $M$ is zero, meaning both have the same horizontal velocity $V$ in the ground frame.

By conservation of horizontal momentum:
$mv = (M + m)V \Rightarrow V = \dfrac{mv}{M + m}$

Let $v_y$ be the vertical velocity of the small body at the moment it breaks off. By conservation of mechanical energy:
$\dfrac{1}{2}mv^2 = \dfrac{1}{2}(M + m)V^2 + \dfrac{1}{2}mv_y^2 + mgh$

Substituting $V$ into the energy equation:
$mv^2 = (M + m)\left(\dfrac{mv}{M + m}\right)^2 + mv_y^2 + 2mgh$

$v^2 = \dfrac{mv^2}{M + m} + v_y^2 + 2gh$

$v_y^2 = v^2 - \dfrac{mv^2}{M + m} - 2gh = \dfrac{Mv^2}{M + m} - 2gh = \dfrac{Mv^2 - 2(M + m)gh}{M + m}$

$v_y = \left[ \dfrac{Mv^2 - 2(M + m)gh}{M + m} \right]^{1/2}$

The time of flight of the small body under gravity is:
$T = \dfrac{2v_y}{g} = \dfrac{2}{g}\left[ \dfrac{Mv^2 - 2(M + m)gh}{M + m} \right]^{1/2}$

During this time, the larger block moves with a constant horizontal velocity $V$. The distance traversed by the larger block is:
$D = VT = \left(\dfrac{mv}{M + m}\right) \dfrac{2}{g}\left[ \dfrac{Mv^2 - 2(M + m)gh}{M + m} \right]^{1/2}$

$D = \dfrac{2mv[Mv^2 - 2(M + m)gh]^{1/2}}{g(M + m)^{3/2}}$