Answer
$2.01~m/s$
Work Step by Step
We know that
$v_f=\sqrt{2g_mh}$
We plug in the known values to obtain:
$v_f=\sqrt{2(1.62m/s^2)(0.95m)}$
This simplifies to:
$v_f=2.01m/s$
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 54: 96
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