Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 13, Trigonometric Ratios and Functions - 13.3 Evaluate Trigonometric Functions of Any Angle - 13.3 Exercises - Problem Solving - Page 871: 36


$16.5 ft/s$

Work Step by Step

Here, we have $\theta=18^{\circ}$ and $d=5$ We know that $d=\dfrac{v^2}{32} \sin (2 \theta)$ $5=\dfrac{v^2}{32} \sin 2 (18^{\circ})$ This implies that $v^2=272.21$ Thus, $v= \sqrt {272.21}=16.5 ft/s$
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