Show that air drag varies quadratically with speed

AI Thread Summary
The discussion focuses on deriving equations that show how speed varies with height when a bullet is fired straight up, considering air drag that varies quadratically with speed. The key equations provided are v^2 = Ae^(-2kx) - g/k for upward motion and V^2 = g/k - Be^(2kx) for downward motion, with constants A and B representing integration constants. The user attempts to relate their derived equations for speed to the original equations but expresses uncertainty about how to connect these concepts effectively. The conversation emphasizes the need to equate the right sides of the acceleration equations to form a differential equation that can be solved for v(x). Ultimately, the goal is to clarify the relationship between speed and height in the context of the problem.
monkeyboy590
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Homework Statement



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.)

Homework 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)

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.
 
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You already know that

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 ...
 
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!
 
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?
 
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