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: 35

Answer

$H_{1}=4.83cm$

Work Step by Step

Lets assume that the initial angle of bullet respect to horizontal line that passes between target and rifle is $\theta$. Since the rifle and the target are located at the same level. So the equation between $\theta$, initial velocity of bullet ($v_{0} $) and the distance between rifle and target (assume the distance is X) will be: $X=\frac{2v_{0}^{2}*sin\theta*cos\theta}{g}$ İf we subsitute $\sin \theta =\dfrac {\tan \theta }{\sqrt {1+\tan ^{2}\theta }}$ and $\cos \theta =\dfrac {1}{\sqrt {1+\tan ^{2}\theta }}$ we get, $X=\dfrac {2v^{2}_{0}}{g}\dfrac {\tan \theta }{\left( 1+\tan ^{2}\theta \right) }$ The distance from aiming point to the target can be defined as $\tan \theta =\dfrac {H}{X}$ So if we substitute $tan\theta$ with $\frac{H}{X}$ we get: $\dfrac {Xg}{2v^{2}_{0}}=\dfrac {\dfrac {H}{X}}{1+\left( \dfrac {H}{X}\right) ^{2}}$ So if we simplify this, $\left( \dfrac {H}{x}\right) ^{2}-\dfrac {2v^{2}_{0}}{xg}\left( \dfrac {H}{x}\right) +1=0$ If we solve the equation we find $H=X\dfrac {\dfrac {2v^{2}_{0}}{Xg}\pm \sqrt {\dfrac {4v^{4}_{0}}{X^{2}g^{2}}-4}}{2}$ İf we put values $v_{0}=460\dfrac {m}{s};X=45,7m;g=9,8\dfrac {m}{s^{2}}$, we get $H_{1}=4,83cm;H_{2}=43,183km$ $H_{2} $ seems impossible so the distance will be $H_{1}=4.83cm$
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