Since the rock is falling we have an uniform accelerated motion. For this kind of motion we know that:
[tex]v_f^2-v_0^2=2a\left(y-y_0\right)[/tex]In this case we have that:
The initial velocity is zero.
The acceleration is 9.8 m/s^2.
The change in position is 100 m.
Plugging this values in the equation we have:
[tex]\begin{gathered} v_f^2-0^2=2\left(9.8\right)\lparen100) \\ v_f^2=1960 \\ v_f=\sqrt{1960} \\ v_f=44.27 \end{gathered}[/tex]Therefore, the final velocity is 44.27 m/s
