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: 70

Answer

60 m/s at 61 degrees.

Work Step by Step

Let the origin be the launch point from which the projectile is launched, and let upward be the positive y direction. Find the horizontal velocity component. $$v_x=\frac{195\;m}{6.6\;s}=29.55m/s$$ Now find the initial y velocity. $$y=y_0+v_{y0}t+0.5a_yt^2$$ $$135m= v_{y0}(6.60s)+0.5(-9.8m/s^2)(6.6s)^2$$ $$v_{y0}=52.79m/s$$ Now that we have the components, find the speed of 60 m/s by using the Pythagorean Theorem. $$v=\sqrt{{v_{0y}^2+}{v_{0x}^2}}\approx 60\;m/s$$ Then find the angle. $$\theta = \tan^{-1}\frac{v_{0y}}{v_{0x}}=61^{\circ}$$
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