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.