Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 3 - Exercises and Problems: 63

Answer

Please see the work below.

Work Step by Step

We know that $t=\sqrt{\frac{2h}{g}}$ We plug in the know values to obtain: $t=\sqrt{\frac{(2)(4.2-1.5)}{9.8}}=0.742s$ The horizontal speed is $v_x=\frac{x}{t}=\frac{3.0}{0.742}=4.04\frac{m}{s}$ The vertical speed is $v_y=gt=9.8(0.742)=7.27\frac{m}{s}$ Now we can find the speed of the package as $v=\sqrt{(v_x)^2+(v_y)^2}=\sqrt{(4.04)^2+(7.27)^2}=8.3\frac{m}{s}$ and the direction of the package is $\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{7.27}{4.04})=61^{\circ}$
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