Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 2 - Exercises and Problems - Page 32: 71

Answer

Please see the work below.

Work Step by Step

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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.