Answer
$17~m/s$
Work Step by Step
Let the positive direction be downward:
$\Delta x=14~m$, $v_0=0$ and $a=g=9.81~m/s^2$
$v^2=v_0^2+2a\Delta x$
$v=\sqrt {2g\Delta x}=\sqrt {2(9.81~m/s^2)(14~m)}=17~m/s$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 2 - One-Dimensional Kinematics - Problems and Conceptual Exercises - Page 53: 74
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