Answer
a). $23.6^{\circ}$
b). $1.46hr$
Work Step by Step
a). $\theta=sin^{-1}\frac{60}{150}=sin^{-1}0.4=23.6^{\circ}$
b). required velocity=$\sqrt (150^{2}-60^{2})=137.5mi/hr$
So, time needed to cover 200 mi $=\frac{200}{137.5}=1.46hr$
College Physics (7th Edition)
by Wilson, Jerry D.; Buffa, Anthony J.; Lou, Bo
Published by
Pearson
ISBN 10:
0-32160-183-1
ISBN 13:
978-0-32160-183-4
Chapter 3 - Motion in Two Dimensions - Learning Path Questions and Exercises - Exercises - Page 102: 84
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