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 3 - Exercises and Problems - Page 52: 77

Answer

$\theta=66^{\circ}$

Work Step by Step

We see that the ball needs to travel 15 feet horizontally and 1.8 feet up. We first convert everything into metric units: $15\ ft = 4.572 m$ $1.8 \ ft = .54864m$ $26 \ ft/s = 7.9248 \ m/s$ We know the following equations: $\Delta x = v_{x0}t$ $\Delta y = v_{y0}t - \frac{1}{2}gt^2$ We use the first equation to find: $4.572=26cos\theta t$ $t =\frac{.1758}{cos\theta}$ We plug this into the second equation to obtain: $\Delta y = 26sin\theta(\frac{.1758}{cos\theta})-\frac{1}{2}(9.81)(\frac{.1758}{cos\theta})^2$ $.54864 = 7.9248sin\theta(\frac{.1758}{cos\theta})-\frac{1}{2}(9.81)(\frac{.1758}{cos\theta})^2$ Solving this equation gives: $\theta=66^{\circ}$
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