College Physics (7th Edition)

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 96: 2

Answer

(a) For $\theta<45^{\circ}$, the horizontal component is greater than the vertical component. (b) For $\theta=37^{\circ}$, the horizontal component is $27.95\,\mathrm{m/s}$ and the vertical component is $21.06\,\mathrm{m/s}$.

Work Step by Step

A velocity $v$ directed at an angle $\theta$ above horizontal has the components - $v_x=v\cos\theta$ $v_y=v\sin\theta$. (a) For $\theta<45^{\circ}$, $\cos\theta>\sin\theta$. Thus, the horizontal velocity component is greater than the vertical velocity component. (b) For $\theta=37^{\circ}$, initial velocity components - $v_x=35\cos37^{\circ}=27.95\,\mathrm{m/s}$ $v_y=35\sin37^{\circ}=21.06\,\mathrm{m/s}$.
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