$A$ mass of $10 \, kg$ is dropped from a height of $50 \, cm$. Find its:
$(a)$ Potential energy just before dropping.
$(b)$ Kinetic energy just on touching the ground.
$(c)$ Velocity with which it hits the ground. (Given: $g = 10 \, m s^{-2}$)

(C) Given: Mass $m = 10 \, kg$,Height $h = 50 \, cm = 0.5 \, m$,Acceleration due to gravity $g = 10 \, m s^{-2}$.
$(a)$ Potential energy $(PE)$ just before dropping is calculated as $PE = mgh = 10 \times 10 \times 0.5 = 50 \, J$.
$(b)$ According to the law of conservation of energy,the total mechanical energy remains constant. Just before touching the ground,the potential energy is converted into kinetic energy $(KE)$. Therefore,$KE = PE = 50 \, J$.
$(c)$ Using the formula for kinetic energy,$KE = \frac{1}{2}mv^2$,we can find the velocity $v$ as $v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 50}{10}} = \sqrt{10} \approx 3.16 \, m s^{-1}$.