# Bullet trajectory when fired at an angle.

1. May 11, 2006

### flyb0y

So I was curious as to how far a high velocity bullet would travel if fired at a 45 deg angle. I used the muzzle velocity for a .306 150 g bullet which is 2700 fps or 822.9 m/s.

I only calculated this for zero air resitance because I quite honestly dont know how to calculate for air resistance. However, I would like some input on if my calculations are right for what I have done and also if somebody could show how much of a difference there would be with air resistance and any other factors calculated.

Used basic kinematic equations for this

initial Y = 1.9m
initial V = 822.97 m/s
theta = 45deg
g = -9.8 m/s^2

Vy = 822.97 sin 45
= 581.9 m/s in Y dir

Vx = 822.97 cos 45
= 581.9 m/s in X dir

Vy = VinitialY + gt
Vy = 0
t= -Vyinitial/-g
t= 59.3 sec
59.3 seconds till Vy = 0 or peak altitude.

ill skip Ymax because im more interested in Xmax
(I got 17,277.9 m for Ymax) ???!?!?!?

Xmax = Xinitial + VXinit(t) + 1/2gt^2
= VXinit(t)
= (581.9)2(59.3)
= 69013.3 m = 226,418.8 ft = 42.9 mi :surprised

42.9 mi horizontal distance traveled???!!!?!?! Can this be anywhere close to correct? That is REALLY far.
I know air resistance will play a role in this distance but just how much?

Thanks for looking guys.

2. May 12, 2006

### Andrew Mason

This is correct. You can just use R = v^2/g if the angle is 45 degrees. ($R = v^2sin2\theta/g$)

Air resistance plays a huge role in determining the range. A bullet will slow down approximately .5 - 1 ft/sec each foot travelled, initially. Also, the longer it travels, the greater the instability of the bullet. It begins to yaw and this makes it less aerodynamic and causes it to slow and change direction. The maximum range for your bullet is probably about a mile at 45 degrees.

AM