Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 3 - Section 3.5 - Projectile Motion - Example - Page 44: 3.5

Answer

5.6 meters

Work Step by Step

First, we set equation 13.4 in the textbook equal to zero to obtain: $(\frac{g}{2(v_0^2)})cos^2{\theta}x^2-tan\theta x + y = 0$ Plugging in the known values, it follows: $x=3.1,8.7$ Since x=3.1 is simply where it crosses the corner, we know we want the second value of x. Thus, we find that the distance it lands from the cliff is: $=8.7-3.1=5.6 \ m$
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