Vertical Motion with Quadratic Air Resistance

AI Thread Summary
The discussion focuses on deriving the equation of motion for a baseball thrown vertically upward, factoring in quadratic air resistance. The initial equation of motion is established as m*v(dot) = -mg - cv², which is rearranged to v(dot) = -g(1 + (v/vter)²) using terminal velocity concepts. The user struggles with separating variables and integrating the resulting equation to express height in terms of velocity. They attempt to integrate but express confusion over the integration process and how to arrive at the maximum height formula involving logarithms. The conversation emphasizes the need for clearer steps in the integration process to solve the problem effectively.
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Before I write anything, I want to apologize because I have no idea how to write equations on this website. This is my first post >.< Also, thank you for helping in advance!

Homework Statement


A baseball is thrown vertically up with speed vo and is subject to a quadratic drag with magnitude f(v) = cv2. Write down the equation of motion for the upward journey (measuring y vertically UP) and show that it can be rewritten as v(dot) = -g[1+(v/vter)2]. Use the "vdv/dx rule" to write v(dot) as vdv/dy and then solve the equation of motion by separating variables (put all terms involving v on one side and all terms involving y on the other). integrate both sides to give y in terms of v, and hence v as a function of y. Show that the baseball's maximum height is

ymax = [(vter)2/2g]*ln[ [ (vter)2 + (vo)2 ] / [(vter)2] ]

whew. If vo = 20m/s and the baseball has the parameters: mass m=.15kg and diameter D = 7cm, what is ymax? Compare with the value in a vacuum.


Homework Equations


Ok... Well first, in case you didn't get it, the vdv/dx rule is just that:

v(dot) = vdv/dx = (1/2)d(v2)/dx.

(only in this problem we just use y instead of x.)

Another formula that's important is the terminal velocity, which is
vter = sqrt(mg/c)


The Attempt at a Solution



Well, the first thing it asks is to write down the equation of motion. I'm a little unsure, but I think that it is :

m*v(dot) = -mg - cv2

which can be rearranged:
v(dot) = -g - cv2/m

and substituting c/m = g/(vter)2 in...
v(dot) = -g (1 + (v/vter)2)

so then we use the vdv/dx rule...
vdv = -g*dy*(1 + (v/vter)2)

and separating variables like it said,
vdv/(1 + (v/vter)2) = -gdy

But now I'm not sure what I'm supposed to do. When it said to separate variables, it said that I should put the terms with a y on one side and the terms with a v on the other, but... are there any terms with a y? Other than the dy? I also have no idea how to integrate this equation... Can anybody help me figure out the next few steps? Thank you again.

PS: is there a way to actually have it write v(dot) normally - as in, with a dot above the v?
 
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Ok so I went ahead and tried to integrate vdv/(1 + (v/vter)2) = -gdy

On the left, I went from vo to v and on the right i went from 0 to y. This gave me (and watch out i switch the left and right sides here):

-gy = sqrt[ (1/vter2)*v2 + 1 ] / (1/vter2) from vo to v.

Now, vo = 0 at ymax, so you can plug those in, and you get

-gymax = (vter)2*sqrt[ (1/vter2)*v2 + 1 ] from vo to 0

which simplifies to

-gymax = vter2 * (1 - sqrt[ (vo/vter)2 +1 ]

and I have no idea how to make that into the given equation (with ln and stuff) that is shown in my first post.
 
Rumble said:
vdv/(1 + (v/vter)2) = -gdy
-gy = sqrt[ (1/vter2)*v2 + 1 ] / (1/vter2)
No, that integration step with v is wrong. Please write it out in more detail.
 
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