Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 2 - Describing Motion: Kinematics in One Dimension - General Problems - Page 47: 77

Answer

The percent difference in the height measurement is 32%.

Work Step by Step

$height = \frac{1}{2}at^2 = \frac{1}{2}(9.8)(t^2)$ $h_1 =\frac{1}{2}(9.8)(2.0)^2 = 19.6 ~m$ $h_2 =\frac{1}{2}(9.8)(2.3)^2 = 25.9 ~m$ The percent difference is $\frac{25.9-19.6}{19.6}\times100\% = 32\%$
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