Answer
Please see the work below.
Work Step by Step
We know that
$v^2=2g_{Mars}d$
This simplifies to:
$v=\sqrt{2g_{Mars}}d$
$v=\sqrt{(2)(3.74)(15)}$
$v=11\frac{m}{s}$
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: 65
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