$A$ man drops a $10 \, kg$ rock from the top of a $20 \, m$ ladder. What will be its kinetic energy when it reaches the ground? What will be its speed just before it hits the ground? (Take $g = 10 \, m s^{-2}$)

(N/A) Given: Mass $m = 10 \, kg$,Height $h = 20 \, m$,Acceleration due to gravity $g = 10 \, m s^{-2}$.
$1$. Kinetic Energy $(KE)$ at the ground:
According to the law of conservation of energy,the potential energy $(PE)$ at the top is converted into kinetic energy $(KE)$ at the ground.
$PE = mgh = 10 \times 10 \times 20 = 2000 \, J$.
Therefore,$KE = 2000 \, J$.
$2$. Speed $(v)$ just before hitting the ground:
Using the formula $KE = \frac{1}{2} mv^2$:
$2000 = \frac{1}{2} \times 10 \times v^2$
$2000 = 5 \times v^2$
$v^2 = 400$
$v = \sqrt{400} = 20 \, m s^{-1}$.