Answer
$\theta = sin^{-1}(2(\frac{gh}{v_0^2}+1))^{-.5}$
Work Step by Step
We know the following equation for the range:
$sin\theta = (2(\frac{gh}{v_0^2}+1))^{-.5}$
$\theta = sin^{-1}(2(\frac{gh}{v_0^2}+1))^{-.5}$
Essential University Physics: Volume 1 (3rd Edition)
by Wolfson, Richard
Published by
Pearson
ISBN 10:
0321993721
ISBN 13:
978-0-32199-372-4
Chapter 3 - Exercises and Problems - Page 50: 85
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