Answer
$v_{f}=42m/s$
Work Step by Step
We know the following equation:
$v^{2}_{f}=v^{2}_{0}+2gx$
We plug in the known values to obtain:
$v_{f}=\sqrt {v^{2}_{0}+2gx}=\sqrt {(0m/s)^{2}+2(9.8m/s^{2})(90m)}=42m/s$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 2 - Motion Along a Straight Line - Problems - Page 35: 52c
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