When mass B (5 kg) is lifted and released, it accelerates downward due to gravity, pulling mass A (5 kg) horizontally on the smooth table. Since the pulley is smooth and massless, the tension is equal on both sides. The acceleration $a$ of the system is given by $a=\frac{mg}{m+m}=\frac{g}{2}$. Mass A moves horizontally with this acceleration. When B hits the ground, A has a velocity $v$. After that, A continues moving with constant velocity (no acceleration) until it rises to a height $h$ on the curved path. Using energy conservation, the kinetic energy of A converts to potential energy: $\frac{1}{2}m{v}^2=mgh$. Solving, $h=\frac{v^2}{2g}$. From kinematics, $v^2=2as$, where $s=1$ m (initial height). Substituting $a=\frac{g}{2}$, $v^2=g$, so $h=\frac{g}{2g}=0.5$ m.

Final answer: 0.5 m