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 73: 68

Answer

$v_{min}=19.4\frac{m}{s}$ $v_{max}=19.5\frac{m}{s}$

Work Step by Step

$v_{fy}=0$ $g=-9.8\frac{m}{s^2}$ From $v_{fy}^2=v_{iy}^2+2a\Delta y$ $v_{iy}=\sqrt{v_{f}^2-2a\Delta y}$ $v_{iy}=\sqrt{0-2(-9.8\frac{m}{s^2})(0.95m)}$ $v_{iy}=18.6\frac{m}{s}$ $t=\frac{v_{fy}-v_{iy}}{a}=\frac{0-18.6\frac{m}{s}}{-9.8\frac{m}{s^2}}=1.9s$ $v_x=\frac{10.78m}{1.9s}=5.67\frac{m}{s}$ $v_x=\frac{11.22m}{1.9s}=5.91\frac{m}{s}$ $v_{min}=\sqrt{\big(18.6\frac{m}{s}\big)^2+\big(5.67\frac{m}{s}\big)^2}=19.4\frac{m}{s}$ $v_{max}=\sqrt{\big(18.6\frac{m}{s}\big)^2+\big(5.91\frac{m}{s}\big)^2}=19.5\frac{m}{s}$
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