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: 60

Answer

William should aim at an angle of $6.24^{\circ}$ above the horizontal.

Work Step by Step

We can use the equation for range $x$ to solve this question: $x = \frac{v_0^2~sin(2\theta)}{g}$ $sin(2\theta) = \frac{xg}{v_0^2}$ $sin(2\theta) = \frac{(27~m)(9.80~m/s^2)}{(35~m/s)^2}$ $sin(2\theta) = 0.216$ $\theta = \frac{arcsin(0.216)}{2}$ $\theta = 6.24^{\circ}$ Therefore, William should aim at an angle of $6.24^{\circ}$ above the horizontal.
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