- #1
mekrob
- 11
- 0
Trajectory of a bullet with drag
In most physics books, they pass over drag problems because of their difficulty. However, I had to do one for a mid-term for a physics class. When the velocity is low (tens of m/s or less), then the drag is proportional to the velocity and the math is quite simple (at least in comparison). However, we were asked about a bullet which travels nearly 1000 m/s and thus is dependent upon velocity squared. I was utterly confused with how to progress with the math.
Here's what I had:
Since the bullet is fired at an initially positive angle [tex]\theta[/tex], the drag can be broken into two components (2-D only): Fx and Fy.
Fx = F * cos[tex]\theta[/tex]
Fy = F * sin[tex]\theta[/tex]
FDrag = [tex]\frac{(Drag Coefficient)*(\rho)*V^{2}*(Area)}{2}[/tex]
FDrag = k*V2, where K = .5*Cd*[tex]\rho[/tex]*Area
Assume that density is constant throughout time.
Fx = m[tex]\frac{\partial vx}{\partial t}[/tex] = -k*V2
[tex]\frac{\partial vx}{\partial t}[/tex] = -k*Vx2/m
[tex]\frac{\partial vx}{V^2}[/tex] =-k/m * dt
Integrate to get:
-1/Vx = -kt/m + C
Vx = m/kt + K
However, at t = 0, Vx = muzzle velocity*cos(theta). But according to the equation I have, Vx would go to infinity as t -> 0. Right?
I run into the same problem when trying to figure out Vy which is even more complicated since it involves gravity. Can anyone tell me where I went wrong with this?
Thanks
In most physics books, they pass over drag problems because of their difficulty. However, I had to do one for a mid-term for a physics class. When the velocity is low (tens of m/s or less), then the drag is proportional to the velocity and the math is quite simple (at least in comparison). However, we were asked about a bullet which travels nearly 1000 m/s and thus is dependent upon velocity squared. I was utterly confused with how to progress with the math.
Here's what I had:
Since the bullet is fired at an initially positive angle [tex]\theta[/tex], the drag can be broken into two components (2-D only): Fx and Fy.
Fx = F * cos[tex]\theta[/tex]
Fy = F * sin[tex]\theta[/tex]
FDrag = [tex]\frac{(Drag Coefficient)*(\rho)*V^{2}*(Area)}{2}[/tex]
FDrag = k*V2, where K = .5*Cd*[tex]\rho[/tex]*Area
Assume that density is constant throughout time.
Fx = m[tex]\frac{\partial vx}{\partial t}[/tex] = -k*V2
[tex]\frac{\partial vx}{\partial t}[/tex] = -k*Vx2/m
[tex]\frac{\partial vx}{V^2}[/tex] =-k/m * dt
Integrate to get:
-1/Vx = -kt/m + C
Vx = m/kt + K
However, at t = 0, Vx = muzzle velocity*cos(theta). But according to the equation I have, Vx would go to infinity as t -> 0. Right?
I run into the same problem when trying to figure out Vy which is even more complicated since it involves gravity. Can anyone tell me where I went wrong with this?
Thanks