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 173: 92

Answer

11.05 Note, since air resistance was considered, the maximum height is lower, causing the value of g to appear higher than it actually is.

Work Step by Step

We know from conservation of energy that $v=\sqrt{2gh}$, and we know that $v=\frac{J}{m}$. (Note, m is mass, g is the gravitational constant, h is the height, and J is the impulse). Setting these equal, we find: $2gh = \frac{J^2}{m^2}$ Thus, we could plot $J^2$ versus $2hm^2$. We could also plot $\frac{J^2}{m^2}$ versus 2h. Doing this, we find the experimental gravitational constant to be: 11.05. Note, since air resistance was considered, the maximum height is lower, causing the value of g to appear higher than it actually is.
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