Show that air drag varies quadratically with speed

1. Sep 24, 2009

monkeyboy590

1. The problem statement, all variables and given/known data

A gun is fired straight up. Assuming that the air drag on the bullet varies quadratically with speed. Show that speed varies with height according to the equations

v^2 = Ae^(-2kx) - g/k (upward motion)
V^2 = g/k - Be^(2kx) (downward motion)

in which A and B are constants of integration, g is the acceleration of gravity, k=cm, where c is the drag constant and m is the mass of the bullet. (Note: x is measured positive upward, and the gravitational force is assumed to be constant.)

2. Relevant equations

m dv/dt = mg - cv^2 = mg(1 - cv^2/mg)
.'. dv/dt = g(1 - v^2/v(sub t)^2) where v(sub t) is the terminal speed, sqrt(mg/c)

dv/dt = dv/dx dx/dt = 1/2 dv^2/dx
So now we can write the equation as:

dv^2/dx = 2g(1 - v^2/v(sub t)^2)

3. The attempt at a solution

I have gotten as far as to say:

For downward motion:
v^2 = g/k(1exp^[2kx]) + v(naught)^2.exp^[2kx]​
Likewise for upward, we get:
v^2 = g/k(1-exp^[-2kx]) + v(naught)^2.exp^[-2kx]​

I just don't know how to relate these to the original equation to show that speed varies with height.

2. Sep 24, 2009

kuruman

$$a=\frac{dv}{dt}=g \large(1-\frac{v^{2}}{v^{2}_{T}} \large)$$

Now it is also true that

$$a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$

So ...

3. Sep 24, 2009

monkeyboy590

To be completely honest, I'm not sure where you're going with that.

I'm not sure how to show specifically that the speed varies with the height. What variable would I use to represent speed, and how do I incorporate the original equations that were given in the problem into this solution?

Thank you so much for your help!

4. Sep 24, 2009

kuruman

Look at the equations I gave you. If the right sides are equal to the same thing, the acceleration, then the right sides are equal to each other. Then you get a differential equation that you can solve to find v(x). Isn't that what you want?