Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 9 - Exercises and Problems - Page 165: 61


5.79 seconds

Work Step by Step

The fragments are equal in mass, so they are also equal in velocity. If this were not the case, momentum would not be conserved. We first must find where the first one explodes: $h = \frac{u^2}{2g}=\frac{40^2}{2(9.81)}=81.55 \ meters$ Using this, we can find the velocity of the first (and thus the second) fragment: $v = \frac{h-\frac{1}{2}gt^2}{t}$ $v = \frac{81.55-\frac{1}{2}(9.81)(2.87)^2}{2.87}=14.34 \ m/s$ Now, we can find t: $h = -v_0t + \frac{1}{2}at^2$ $0 =-h -v_0t + \frac{1}{2}at^2$ Simplifying this using the quadratic formula, it follows: $t = \frac{2.923\pm\sqrt{2.923^2-4(1)(-16.64)}}{2(1)}=5.79 \ s$
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.