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Vertical Motion with Quadratic Air Resistance

  1. Jan 24, 2013 #1
    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!

    1. The problem statement, all variables and given/known data
    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.


    2. Relevant 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)


    3. 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?
     
  2. jcsd
  3. Jan 24, 2013 #2
    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.
     
  4. Jan 24, 2013 #3

    haruspex

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    No, that integration step with v is wrong. Please write it out in more detail.
     
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