Answer
1800 ft
Work Step by Step
Magnitude of the horizontal component of the velocity vector in problem 32 is $|\textbf{V}_{x}|=900\,ft/s$. Given that $t= 2\,s$.
Horizontal distance=|$\textbf{V}_{x}$|$\times$time$=900\,ft/s\times2\,s=1800\,ft$
Trigonometry 7th Edition
by McKeague, Charles P.; Turner, Mark D.
Published by
Cengage Learning
ISBN 10:
1111826854
ISBN 13:
978-1-11182-685-7
Chapter 2 - Section 2.5 - Vectors: A Geometric Approach - 2.5 Problem Set - Page 107: 34
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