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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

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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

Jul 17, 2015

#2
+27908
+5

"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!

.

Jul 25, 2015

#1
+101745
+5

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.

Jul 17, 2015
#2
+27908
+5

"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!

.

Alan Jul 25, 2015
#3
+101745
0

Thanks Alan that makes good sense :)

Jul 25, 2015