First of all, let us convert the given data into the standard units.
Velocity = 1790 mi/h

The bullet was fired from a height of 5 ft.

Now, we need to find out the time it takes for the bullet to reach the ground.

Where g is the acceleration due to gravity and u is the initial velocity of the bullet which must be zero since the bull was at rest initially.
![\begin{gathered} s=u\cdot t+(1)/(2)gt^2 \\ 1.524=0\cdot t+(1)/(2)\cdot9.81\cdot t^2 \\ 1.524=(1)/(2)\cdot9.81\cdot t^2 \\ t^2=(2\cdot1.524)/(9.81) \\ t^2=0.311 \\ t=\sqrt[]{0.311} \\ t=0.558\; s \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/uwb1lm6yhzo9az63sy7gxccu03uqvj6zw4.png)
The horizontal distance covered by the bullet is given by

Therefore, the range of the bullet is 446.4 m
Or 1464.6 ft