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: 76

Answer

Please see the work below.

Work Step by Step

We know that the range of the first projectile is given as: $R_1=\frac{v_{\circ}^2 sin(2\theta)}{g}$ $R_1=\frac{v_{\circ}^2 sin(2(45+\alpha))}{g}=\frac{v_{\circ}^2 sin(90+2\alpha)}{g}$ We can find the range of the second projectile as: $R_2=\frac{v_{\circ}^2 sin(2\theta)}{g}$ $R_2=\frac{v_{\circ}^2 sin(2(45-\alpha))}{g}=\frac{v_{\circ}^2 sin(90-2\alpha)}{g}$ As we know that $sin(90+\theta)=sin(90-\theta)$, the range of the projectile is the same.
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