Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.2 - Integrals of Vector Functions; Projectile Motion - Exercises 13.2 - Page 753: 22


$$ 55.4 \mathrm{ft}$$

Work Step by Step

Since $$y=y_{0}+\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2}$$ Then $$ y=32+16 t-16 t^{2}$$ The ball hits the ground when $y=0 $ $$\Rightarrow 0=32+16 t-16 t^{2} $$ Hence $ t=-1$ or $t=2 t=2 \sec$ since $t>0 ;$ thus,\begin{align*} x&=\left(v_{0} \cos \alpha\right) t\\ x&=(32 \mathrm{ft} / \mathrm{sec})\left(\cos 30^{\circ}\right) t\\ &=32\left(\frac{\sqrt{3}}{2}\right)(2) \approx 55.4 \mathrm{ft} \end{align*}
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