$A$ man drops a $10\, kg$ rock from the top of a $5\, m$ ladder. What is its kinetic energy when it reaches the ground? What is its speed just before it hits the ground?

(N/A) Given: Mass,$m = 10\, kg$,Height,$h = 5\, m$,Acceleration due to gravity,$g = 9.8\, m/s^2$.
$1$. Potential Energy $(PE)$ at the top:
$PE = mgh = 10 \times 9.8 \times 5 = 490\, J$.
$2$. Kinetic Energy $(KE)$ at the ground:
According to the law of conservation of energy,the potential energy at the top is converted into kinetic energy at the ground.
$KE = PE = 490\, J$.
$3$. Speed just before hitting the ground:
Using the equation of motion $v^2 - u^2 = 2gh$,where initial velocity $u = 0\, m/s$:
$v^2 - 0 = 2 \times 9.8 \times 5$
$v^2 = 98$
$v = \sqrt{98} \approx 9.9\, m/s$.