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

Guest Jul 17, 2015

#2**+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 velocity^{n}, 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

#1**+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.

Melody
Jul 17, 2015

#2**+5 **

Best Answer

"*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 velocity^{n}, 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