How far will a .223 travel when shot at a 45 degree angle

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TwoWheelFiend

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just wondering. I should of listened to my math teachers when they told me id need math later in life.
anyone have any idea?
 
First you find the time in flight:
vf=vi-at
0=940sin(45)-9.8t (assuming SS109 projectile from a 20 in barrel and earths gravity)
t=67.8, but this is only from when the bullet leaves the barrel to the apex of the curve so:
t=135seconds. TWO MINUTES of flight time.

Then assuming no wind drag you do distance=time*velocity:
d=940cos(45)m/s*135s
d=45078m
d=45 Kilometers. ~30 mi
Granted its been a while since I took Physics :D Correct my math if I'm wrong.
 
this is what ive always heard...the 223-5.56mm max range is 3,600 meters, the maximum effective range is 300 meters. dont know if it is accurate info or not actually.
 
Maximum (effective) range is different than maximum distance. The trajectory of a bullet must be factored into any calculations.

With the small-arms bullet, the effect of air resistance is so great that the maximum distance is obtained at an elevation of about 29°. There is little change as the gun is elevated to about 35°. At angles greater than 35° the bullet will begin to fall closer to the shooter.

If you can follow the math see this - http://www.worsleyschool.net/science/files/bullet/trajectory.html

Texas Park and Wildlife has a chart of distances bullets can travel for their hunter education safety course.

http://www.tpwd.state.tx.us/learning/hunter_education/homestudy/firearms/bullets.phtml



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