#### 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 $H_{1}=4.83cm$