Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 3 - Section 3.5 - Projectile Motion - Example - Page 45: 3.6

Answer

It can be launched at $90^{\circ}-.66^{\circ}=89.33^{\circ}$.

Work Step by Step

We call $\theta$ the amount below 90 degrees that the rocket can fire. Thus, we find: $x = \frac{v_0^2}{g}sin(2\theta)$ $50,000 = \frac{4600^2}{9.81}sin(2\theta)$ $.0231 = sin(2\theta)$ $2\theta = sin^{-1}(.0231)$ $\theta =.66^{\circ}$
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