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- Thread starter koko122
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The graph shows the following:

$$ {v_{\textrm{T}}}^{2} \propto m$$

and from this I want to show that:

$$\Sigma F=mg-D{v_{\textrm{T}}}^{2}=0$$

So that the drag force can be defined as following:

$$F_{drag}=Dv^{2}$$

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mg = .5C

Your graph is for

v

The standard equation for a straight line is..

y = slope*x + constant

but constant = 0

So in this case

y = v

x = mass m

and the slope of the graph

= g/.5C

You know the slope.

You know the area A

You know g

That just leaves C

Sorry had to edit a few typos in this post.

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I don't see how that would follow using only the graph. It follows from Newton's second law, and that ∑That's obvious but from the graph that I have provided can I derive that mg=dv^2 where d is the drag constant? Or can I show that the proportionality constant is equal to g/d somehow?

The graph shows the following:

$$ {v_{\textrm{T}}}^{2} \propto m$$

and from this I want to show that:

$$\Sigma F=mg-D{v_{\textrm{T}}}^{2}=0$$

See C. Watters post #6.

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D = .5C

Had to edit for typo again.

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And wouldn't C

D = .5C_{p}A

Had to edit for typo again.

From http://en.wikipedia.org/wiki/Drag_(physics)#Drag_at_high_velocity

we have

[tex]F_{drag}=\frac{1}{2} \ \rho_{air} \ C_d \ v^2[/tex]

where

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