Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 4 - Motion in Two and Three Dimensions - Problems: 25

Answer

43 m/s

Work Step by Step

We are given the horizontal range of the jump was 77m. We are also told that the angle to the horizontal is $12^{\circ}$ We use the equation for range: $R = \frac{v_{0}^2}{g}sin2\theta_{0} $ This becomes $77 = \frac{v_{0}^2}{9.80}sin(2(12^{\circ})) $ and solving for $v_{0}$ we find that it is 43 m/s
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