Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 4 - Two-Dimensional Kinematics - Problems and Conceptual Exercises - Page 107: 54

Answer

$46.2^{\circ}$

Work Step by Step

At the highest point, we know that the final velocity becomes zero: $\implies v_y=0$ $v_{\circ}sin\theta -gt=0$ This simplifies to: $\theta=sin^{-1}(\frac{gt}{v_{\circ}})$ We plug in the known values to obtain: $\theta=sin^{-1}(\frac{9.81(0.750)}{10.2})$ $\theta=46.2^{\circ}$
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