No

Range, $R=3 \,km$ Angle of projection, $\theta=30^{\circ}$ Acceleration due to gravity, $g=9.8 \,m / s ^{2}$

Horizontal range for the projection velocity $u_{0}$, is given by the relation:

$R=\frac{u_{0}^{2} \sin 2 \theta}{g}$

$3=\frac{u_{0}^{2}}{g} \sin 60^{\circ}$

$\frac{u_{0}^{2}}{g}=2 \sqrt{3}$

The maximum range $\left(R_{\max }\right)$ is achieved by the bullet when it is fired at an angle of $45^{\circ}$ with the horizontal, that is, $R_{\max }=\frac{u_{0}^{2}}{g}$

On comparing above equations , we get:

$R_{\max }=3 \sqrt{3}=2 \times 1.732=3.46 \,km$

Hence, the bullet will not hit a target $5 \,km$ away.