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 9 - Exercises and Problems - Page 172: 65

Answer

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: $t = \frac{2.923\pm\sqrt{2.923^2-4(1)(-16.64)}}{2(1)}=5.79 \ s$
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