Physics: Principles with Applications (7th Edition)

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

Chapter 3 - Kinematics in Two Dimensions; Vectors - General Problems: 58

Answer

a. 8.9m/s b. 10m

Work Step by Step

a. Use the level horizontal range formula to find the takeoff speed of v = 8.854 m/s. The range, takeoff angle, and g are known. b. Let the end point be at y=0, and choose upward to be positive. The starting point is at y = 2.5 m, and the vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. Use equation 2.11b to solve for the time to fall a distance of 2.5 m below the starting height. The initial vertical velocity is $(8.854 m/s)(sin 45 ^{\circ})$. The time, which must be positive, is 1.597 seconds. Now find the horizontal displacement during the jump. $(8.854 m/s)(cos 45 ^{\circ})(1.597 s) = 10.0 m.$ She lands right where she is supposed to.
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