Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 4 - Two-Dimensional Kinematics - Problems and Conceptual Exercises - Page 106: 33

Answer

The solution is $$l=1.26\text{ m}.$$

Work Step by Step

The ball has to cover $h=0.80\text{ m}$ vertical distance moving with the initial speed of $v_{y0}=4.3\sin15^\circ\text{ m/s}$ which means that the final vertical speed is $$v_{yf}=\sqrt{v_{y0}^2+2gh}$$ which gives for total time $$t=\frac{v_{yf}-v_{y0}}{g}=\frac{\sqrt{v_{y0}^2+2gh}-v_{y0}}{g}=\\ \frac{\sqrt{(4.3\cdot\sin15^\circ)^2+2\cdot9.81\cdot0.80}-4.3\sin15^\circ}{9.81}\text{ s}=0.30\text{ s}.$$ Since the ball moves horizontally with the constant component of the velocity, the distance covered is $$l=4.3\text{ m/s }\cdot\cos15^\circ\cdot t=1.26\text{ m}.$$
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