i need to know how far a .50 cal round will go if fired at a 45 degree angle with a muzzle velocity of 2850fps

"48 miles.    Mmmm Now that sounds like a stretch."

In reality there will be some air-drag resistance.  The resistive force tends to be proportional to velocityⁿ, and while the bullet is supersonic n can be as high as 6.  This would have a significant impact on the total distance travelled!

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i need to know how far a .50 cal round will go if fired at a 45 degree angle with a muzzle velocity of 2850fps

The initial vertical velocity will be 2850*sin(45)=1425*sqrt(2)  f/s

The initial horizontal velocity will be 2850*cos(45)=1425*sqrt(2)  f/s

Let the initial point be x=0,y=0

$$\\\ddot y=-32\;fps^2\\\\
\dot y=-32t+1425\sqrt2\;fps\\\\
y=-16t^2+1425\sqrt2t+0\;f\\\\
$Find t when y=0$\\\\
0=-16t^2+1425\sqrt2t\\\\
0=t(-16t+1425\sqrt2)\\\\
t=0\qquad or \qquad 16t=1425\sqrt2\\\\
t=0\qquad or \qquad t=\frac{1425\sqrt2}{16}sec\\\\\\$$

$$\\\ddot x=0\\\\
\dot x=1425\sqrt2\\\\
x=1425\sqrt2 \;t\;\;\; feet\\\\\\
When\;\;t=\frac{1425\sqrt2}{16}sec\\\\
x=1425\sqrt2 \times\frac{1425\sqrt2}{16} \;\;\; feet\\\\
x=\frac{1425^2*2 }{16} \;\;\; feet\\\\$$

$${\frac{{{\mathtt{1\,425}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{16}}}} = {\mathtt{253\,828.125}}$$

253828 feet  

$${\frac{{\mathtt{253\,828}}}{{\mathtt{5\,280}}}} = {\mathtt{48.073\: \!484\: \!848\: \!484\: \!848\: \!5}}$$

48 miles.    Mmmm Now that sounds like a stretch.   

Thanks Alan that makes good sense :)