Bullet trajectory when fired at an angle.

[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 don't 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 I am 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

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.

 

[Andrew Mason]

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?

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

 

[kyle8921]

I would first like to commend you for your curiosity and for taking the initiative to calculate the trajectory of a high velocity bullet. Your calculations seem to be correct based on the information provided.

However, as you mentioned, air resistance is a major factor that affects the trajectory of a bullet. In order to accurately calculate the distance traveled, you would need to take into account the drag force caused by air resistance. This can be quite complex and requires advanced mathematical equations.

Additionally, other factors such as the bullet's shape, weight, and aerodynamics also play a role in its trajectory. Therefore, it is difficult to accurately predict the distance traveled without taking all of these factors into account.

In terms of your calculated distance of 42.9 miles, it is possible for a high velocity bullet to travel this far under ideal conditions, but it is not a common occurrence. Realistically, the distance traveled would be much shorter due to the aforementioned factors.

I would suggest consulting with a ballistic expert or using specialized software to accurately calculate the trajectory of a bullet taking into account all of the relevant factors. This will provide a more accurate estimation of the distance traveled at a 45 degree angle.

Overall, your curiosity and calculations demonstrate a scientific mindset, but it is important to consider all factors and variables to obtain accurate results in any scientific inquiry.