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 - Exercises and Problems - Page 52: 85

Answer

19 meters

Work Step by Step

We know the following equation: $R=\frac{v_0^2sin(2\theta)}{g}$ We find the initial velocity: $v_0=\sqrt{\frac{Rg}{sin(2\theta)}}$ $v_0=\sqrt{\frac{(28)(9.81)}{sin(2(40^{\circ}))}}=16.7\ m/s$ We know the following equations: $\Delta x = v_{0x}t = 16.7cos(40)t$ $\Delta y = v_{0y}t-\frac{1}{2}gt^2=16.7sin(40)t-4.9t^2$ We know: $tan\theta = \frac{opposite}{adjacent}$ Thus, we find: $tan15=\frac{12.79t}{10.73t-4.9t^2}$ $t=1.485 \ s$ We multiply this by the equation for x to find: $x = 16.7cos(40)t=16.7cos(40)(1.485)=19 \ meters$
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