Answer
$v_{f}=\sqrt {v^{2}_{0}+2gh}$
Work Step by Step
This formula is in the reference table ($a$ is replaced with $g$ because acceleration is due to gravity): $v_{f}^{2}={v^{2}_{0}+2gh}$
Algebraically solve for $v_{f}$ : $v_{f}=\sqrt {v^{2}_{0}+2gh}$
