Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 4 - Motion in Two and Three Dimensions - Problems: 40b

Answer

horizontal distance at 45 degrees is less than that at 42 degrees.

Work Step by Step

Lets first calculate the time it passes till the shot hits the ground. We can use $-H=v_{0}t\sin \alpha -g\dfrac {t^{2}}{2}$ to calculate this time. Here, H is initial height of shot, $v_{0}$ is initial velocity and $α$ is initial angle of velocity with respect to the horizontal line. The minus sign indicates that the shot will travel in the (-) y direction with the magnitute of H distance in order to hit the ground. Lets solve this equation: $-2160m=15\dfrac {m}{s}\times \sin 42^{0}\times t-\dfrac {9.8\dfrac {m}{s^{2}}t^{2}}{2}$ After solving, we find $t=22.0448s$. For the horizontal distance traveled by object : $L=v_{0}t\cos \alpha $ So we get, $L=15\dfrac {m}{s}\times \cos 42^{0}\times 22.0448s=245.737m$ Therefore, $L_{\alpha =45}=234.47m$ And $L_{\alpha =42}=245.737m$ We prove that $L_{\alpha =42} > L_{\alpha =45}$ so $α=45^{0}$ doesnt maximize the horizontal distance when launching and the landing height is also different.
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