Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 2 - Motion Along a Straight Line - Problems - Page 38: 97a

Answer

The elevator strikes the ground with a speed of $~~48.5~m/s$

Work Step by Step

We can let "down" be the positive direction. We can find the speed when the elevator strikes the ground: $v^2 = v_0^2+2ay$ $v^2 = 0+2ay$ $v = \sqrt{2ay}$ $v = \sqrt{(2)(9.8~m/s^2)(120~m)}$ $v = 48.5~m/s$ The elevator strikes the ground with a speed of $~~48.5~m/s$
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.