In a gravity-free space, there is no external force acting on the system consisting of the man and the stone. Therefore, the total momentum of the system is conserved. Initial momentum of the system = $0$ (since both are at rest). Let the velocity of the stone be $v_s$ and the velocity of the man be $v_m$. Taking the upward direction as positive, $v_s = -2 \text{ m s}^{-1}$. According to the law of conservation of momentum: $m_m v_m + m_s v_s = 0$ $50 \times v_m + 0.5 \times (-2) = 0$ $50 v_m - 1 = 0$ $v_m = \frac{1}{50} = 0.02 \text{ m s}^{-1}$ So, the man moves upwards with a constant speed of $0.02 \text{ m s}^{-1}$. The time taken by the stone to reach the floor (a distance of $10 \text{ m}$ downwards) is: $t = \frac{\text{Distance}}{\text{Speed}} = \frac{10 \text{ m}}{2 \text{ m s}^{-1}} = 5 \text{ s}$ In this time of $5 \text{ s}$, the upward distance moved by the man is: $d_m = v_m \times t = 0.02 \text{ m s}^{-1} \times 5 \text{ s} = 0.1 \text{ m}$ The final distance of the man above the floor is his initial height plus the distance moved upwards: Final height = $10 \text{ m} + 0.1 \text{ m} = 10.1 \text{ m}$.