# Homework Help: 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?