Bullet Range Neglecting Air Drag

[nikefan5555]

Hello All,

Homework Statement

I was at the firing range the other day when the gun instructor mentioned that the bullet of a AR-15 rifle can travel 5 - 6 miles depending on the angle of the gun when the bullet leaves the muzzle. I wanted to confirm this for myself so, I tried to find the range by plugging in all the known values to the parabolic trajectory equation.

Neglecting air drag and taking only gravity into account, how far does the bullet travel? Here are the know values:

Muzzle Velocity = 975m/s
Height of gun = 1.7m
θ = 45°

Homework Equations

y = (Tan θ)x_o - (g/(2v²_oCos²θ))x²
y = Height of gun [m]. This is -1.7m in this instance
v_o = Initial velocity/muzzle velocity [m/s]
x = unknown distance [m]

The Attempt at a Solution

When I plug all the values into the equation, I get the quadratic equation:

0 = 1.7+1x-.00001x²

Solving the equation gives me x = -1.69997 and x = 100,002m

This is clearly over 5 - 6 miles. This is about 61 miles. I tried this several times, and I am stumped. My questions are:

• Should I use another formula?
• Does ignoring air drag cause the discrepancy in the calculations?

If someone can steer me in the right direction, I can do the calculations myself. Any help would be appreciated.

Thanks!

 

[barryj]

First calculate the time the bullet is in the air b y knowing the initial vertical velocity.
Then, once you have the time, calculate the distance it will travel horizontally by knowing the horizontal velocity.

Ignoring drag, I get the same answer as you do.

 

[barryj]

As you might already know, 45 degrees is the optimum angle to get the greatest distance if there is no drag or other influences. The rifle used normally would be fired mostly horizontally and the range would be substantially lower.

 

[nikefan5555]

Thanks for the quick response! Yes, I calculated the time the bullet is aloft. However, I still get the same answer for the range. Is 61 miles a reasonable answer for a bullet to travel when aimed at 45 degrees? Something seems wrong with this answer.

I also calculated the range when the rifle is fired horizontally. I get 574 m (626 yards). The effective range of the rifle is roughly 400 yards so, 574m seems like a reasonable answer.

 

[nikefan5555]

Sorry. Please let me clarify. I am looking for the distance of the bullet from the muzzle until it hits the ground.

 

[barryj]

YOu are doing the problem correctly I think. The time in the air, at 45 degrees, is 143 seconds. This time multiplied by the horizontral velocity gives you about 100,000 and this is correct. If you fire the rifle exactly horizontal, the time to hit the ground depends on how high the rifle is above ground when fired, in this case 1.7 meters. You can calculate the time to fall 1.7 meters and then multiply this by the rifle bullet speed of 975 and get the range.

 

[barryj]

574 is correct when fired horizontally.

 

[nikefan5555]

Yes. I get 143 seconds in the air as well. What has me stumped is the fact that a bullet without drag goes 61 miles at a 45 degree angle. This means that I can shoot a bullet from the middle of Long Island, NY and it will land in Montauk (tip of Long Island). Does drag really make that much of difference when calculating distance of a projectile?

 

[barryj]

Yes! Drag is a big factor. The retarding force due to drag is proportional to the velocity squared. As it turns out, to get the maximum distance from a projectile, you have to shoot at an angle of about 30 degrees as I recall. Projectiles have pointed shapes to reduce drag but it is still a major factor.

You just can't break the laws of physics.