Answer
Please see the work below.
Work Step by Step
We know that
$\Delta d=\frac{v^2-v_{\circ}^2}{2a} $
We plug in the known values to obtain:
$\Delta d=\frac{(3)^2}{2(9.8)}$
$\Delta d=0.46m$
Essential University Physics: Volume 1 (3rd Edition)
by Wolfson, Richard
Published by
Pearson
ISBN 10:
0321993721
ISBN 13:
978-0-32199-372-4
Chapter 2 - Exercises and Problems - Page 30: 54
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