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: 84

Answer

$\theta=34.38^{\circ}$

Work Step by Step

We know the following equation for the range: $\theta = sin^{-1}(2(\frac{gh}{v_0^2}+1))^{-.5}$ $\theta = sin^{-1}(2(\frac{g(75)}{36^2}+1))^{-.5}$ $\theta=34.38^{\circ}$
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