## Fundamentals of Physics Extended (10th Edition)

$v_{f}=37.6m/s$
Recall the following equation: $v^{2}_{f}=v^{2}_{0}+2gx$ We plug in the known values to obtain: $v_{f}=\sqrt {v^{2}_{0}+2gx}=\sqrt {(0m/s)^{2}+2(9.8m/s^{2}(90m)(.8)}=37.6m/s$