Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 9 - Exercises and Problems - Page 172: 63

Answer

a) 37.7 degrees b) $-.657\ m/s$

Work Step by Step

a) This is a matter of kinematics. We first use the equation for change in y: $\Delta y = v_{0y}t + \frac{1}{2}gt^2$ $0 = v_{0y}t + \frac{1}{2}gt^2$ $-v_{0y}t = \frac{1}{2}gt^2$ $v_{0y} = -\frac{1}{2}gt$ $t = \frac{-2v_0 y}{g}$ We also find: $\Delta x = v_{0x}t$ Thus, it follows: $15.2 = v_{0x}\frac{-2v_0 y}{g}$ $74.56=v_{0x}v_{0y}$ $74.56=v_{0}^2sin\theta cos\theta$ $.51775=sin\theta cos\theta$ $\theta=37.7^{\circ}$ b) We use conservation of momentum: $0 = (12)(4.5)cos(37.7^{\circ})+65v\\ v=-.657\ m/s$
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