Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 66: 47


The flea goes up to a height of 5.15 cm

Work Step by Step

We can find the velocity $v$ after the flea accelerates through the first 0.50 mm. $v^2= v_0^2+2ay = 0 + 2ay$ $v = \sqrt{2ay} = \sqrt{(2)(1000~m/s^2)(0.50\times 10^{-3}~m)}$ $v = 1.0~m/s$ We can find the distance the flea goes up after the initial period of acceleration. $y = \frac{0-v^2}{2a} = \frac{-(1.0~m/s)^2}{(2)(-9.80~m/s^2)}$ $y = 0.0510~m$ The total height is 0.0510 m plus the initial 0.50 mm while the flea was accelerating upward. Thus, the total height is 0.0515 meters which is 5.15 cm.
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