Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 2 - Exercises and Problems: 65

Answer

Please see the work below.

Work Step by Step

We know that $v^2=2g_{Mars}d$ This simplifies to: $v=\sqrt{2g_{Mars}}d$ $v=\sqrt{(2)(3.74)(15)}$ $v=11\frac{m}{s}$
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