As illustrated in the figure, the track is frictionless. Block B of...

Let $v_0$ be the initial velocity of block A.

By conservation of mechanical energy, the velocity $v_1$ of block A just before colliding with block B is given by:
$\dfrac{1}{2}mv_0^2 + mgh = \dfrac{1}{2}mv_1^2$
$\Rightarrow v_1^2 = v_0^2 + 2gh$

During the perfectly elastic collision between block A (mass $m$) and block B (mass $2m$, initially at rest), the velocity $v_B$ of block B just after the collision is:
$v_B = \dfrac{2m}{m + 2m}v_1 = \dfrac{2}{3}v_1$

For block B to reach the sleeping man at height $h$, its kinetic energy just after the collision must be greater than its potential energy at height $h$:
$\dfrac{1}{2}(2m)v_B^2 > (2m)gh$
$\Rightarrow v_B^2 > 2gh$

Substituting $v_B = \dfrac{2}{3}v_1$:
\left(\dfrac{2}{3}v_1\right)^2 > 2gh$<br>$\Rightarrow \dfrac{4}{9}v_1^2 > 2gh$<br>$\Rightarrow v_1^2 > \dfrac{18}{4}gh = 4.5gh$<br><br>Now, substituting$v_1^2 = v_0^2 + 2gh$:<br>$v_0^2 + 2gh > 4.5gh$<br>$\Rightarrow v_0^2 > 2.5gh$<br>$\Rightarrow v_0 > \sqrt{2.5gh}$<br><br>Thus, the initial velocity imparted to block A must be greater than$\sqrt{2.5gh}$.$