- #1

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derivation.

I take the drag force as being [tex] D \ = \ \frac{1}{4} \ A \ v^2 [/tex].

I want to imagine dropping an object from the highest point at which the equation w = mg

will be a satisfactory approximation for the weight.

My problem is understanding how the process works, imagine dropping the object straight

down from the highest point and we set up a free-body diagram as so:

[tex] w_{weight} \ - \ f_{drag} \ = \ ma[/tex]

[tex] mg \ - \ \frac{1}{4} \ A \ v^2 \ = \ m\frac{dv}{dt}[/tex]

[tex] \ m\frac{dv}{dt} \ = \ mg \ - \ \frac{1}{4} \ A \ v^2 [/tex]

[tex] \ \frac{dv}{dt} \ = \ g \ - \ \frac{1}{4m} \ A \ v^2 [/tex]

[tex] \ dv \ = \ ( \ g \ - \ \frac{1}{4m} \ A \ v^2 \ ) \ dt [/tex]

[tex] \ \int_{v_initial}^{v_{final}} \ dv \ = \ \int_{v_initial}^{v_{final}} \ ( \ g \ - \ \frac{1}{4m} \ A \ v^2 \ ) \ dt [/tex]

[tex]v_{final} = v_{initial} \ + \ ( \ gt \ - \ \frac{1}{12m} \ A \ v^3 \ ) |^{v_{final}}_{v_{initial}} [/tex]

I'm thinking that may be wrong, unless the velocity is a function of time which I don't

think it is in the set up I've given, right? I should have done:

[tex]v_{final} = v_{initial} \ + \ ( \ gt \ - \ \frac{1}{4m} \ A \ v^2 \ t \ ) |^{final}_{initial} [/tex]

I think so!

Anyway, my other concern is with the terminal speed, when the forces balance I can write:

[tex] w_{weight} \ - \ f_{drag} \ = \ m \ \cdot \ 0[/tex]

[tex] mg \ - \ \frac{1}{4} \ A \ v^2 \ = \ 0 [/tex]

[tex] mg \ = \ \frac{1}{4} \ A \ v^2 [/tex]

[tex] v \ = \ \sqrt{ \frac{4mg}{A} } [/tex]

Going off what I've done so far, which has the very real potential of being wrong, I

could write that the velocity I just calculated should also equal the integral I did

above seeing as they both give the velocity you'll be falling with from this point on.

[tex] \sqrt{ \frac{4mg}{A} } \ = \ v_{initial} \ + \ ( \ gt \ - \ \frac{1}{4m} \ A \ v^2 \ t \ ) |^{v_{final}}_{v_{initial}} [/tex]

The initial velocity is zero in this situation.

I'm just thinking, what is the set up, You do an integral to find say the height fallen

from the initial point to the place where terminal velocity kicks in and then you can use

the equations of constant acceleration to find the time/height etc... from the ground.

Just trying to get it all together, any help would be greatly appreciated!