Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 3 - Exercises and Problems: 81

Answer

Please see the work below.

Work Step by Step

We know that $x=\frac{v_{\circ}^2}{g} sin (2\theta)$ After differentiating, we obtain: $x^{\prime}=(\frac{v_{\circ}^2}{g})(-2cos(2\theta))$ as $x^{\prime}=0$ $\implies cos(2\theta)=0$ $2\theta=90$ $\theta=45^{\circ}$
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