Hello!
When the rock is dropped, it only contains Gravitational Potential Energy, and when it hits the ground, it contains Kinetic Energy.
So:
PE = KE
[tex]\large\boxed{mgh = \frac{1}{2}mv^2}[/tex]
We can rearrange to solve for velocity. Cancel out mass and solve.
[tex]gh = \frac{1}{2}v^2\\\\2gh = v^2\\\\v = \sqrt{2gh}[/tex]
Plug in the givens:
[tex]v = \sqrt{2(9.8)(4)} = \boxed{8.85 \frac{m}{s}}[/tex]
