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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): F_{x}and F_{y}.

F_{x}=F* cos[tex]\theta[/tex]

F_{y}=F* sin[tex]\theta[/tex]

F_{Drag}= [tex]\frac{(Drag Coefficient)*(\rho)*V^{2}*(Area)}{2}[/tex]

F_{Drag}= k*V^{2}, where K = .5*C_{d}*[tex]\rho[/tex]*Area

Assume that density is constant throughout time.

F_{x}= m[tex]\frac{\partial vx}{\partial t}[/tex] = -k*V^{2}

[tex]\frac{\partial vx}{\partial t}[/tex] = -k*V_{x}^{2}/m

[tex]\frac{\partial vx}{V^2}[/tex] =-k/m * dt

Integrate to get:

-1/V_{x}= -kt/m + C

V_{x}= m/kt + K

However, at t = 0, V_{x}= muzzle velocity*cos(theta). But according to the equation I have, V_{x}would go to infinity as t -> 0. Right?

I run into the same problem when trying to figure out V_{y}which is even more complicated since it involves gravity. Can anyone tell me where I went wrong with this?

Thanks

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# Trajectory of a bullet with drag

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