Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 3 - Kinematics in Two Dimensions; Vectors - General Problems - Page 72: 59

Answer

a) $9.96 $seconds b) $530m$ c) $53.2\frac{m}{s}; -60.3\frac{m}{s}$ d) $80.4\frac{m}{s}$ e) $48.6^o$ below the horizon f) $70.9 m$

Work Step by Step

a) This is in the vertical direction with up positive, down negative. $\Delta x_f=\Delta x_i+v_{0y}t+\frac{1}{2}gt^2$ $v_{0y}=65.0\frac{m}{s}\times \sin(35.0^o)=37.3\frac{m}{s}$ $-115m=0m+37.3\frac{m}{s}t+\frac{1}{2}(-9.8\frac{m}{s^2})t^2$ $-4.9t^2+37.3\frac{m}{s}+115=0$ $t=9.96s$ b) $x=v_{0x}t$ $v_{0x}=65.0\frac{m}{s}\times \cos(35.0^o)=53.2\frac{m}{s}$ $x=(53.2\frac{m}{s})(9.96s)=530$ c) $v_{fy}=v_{0y}+gt$ $v_{fy}=37.3\frac{m}{s}+(-9.8\frac{m}{s})\times9.96s=-60.3\frac{m}{s}$ The horizontal component of the velocity is constant $v_{fy}=-53.2\frac{m}{s}$ d) $v=\sqrt{\big(53.2\frac{m}{s}\big)^2+\big(-60.3\frac{m}{s}\big)^2}=80.4{m}{s}$ e) $\arctan\Big(\frac{-60.3\frac{m}{s}}{53.2\frac{m}{s}}\Big)=48.6^o$ f) $(v_{fy})^2=(v_{0y})^2+2g\Delta y$ $(0\frac{m}{s})^2=(37.3\frac{m}{s})^2+2(-9.8\frac{m}{s^2})\Delta y$ $\Delta y_{max}=\frac{\big(37.3\frac{m}{s}\big)^2}{2\times9.8\frac{m}{s^2}}=70.9m$
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