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Homework Help: Drag force (air resistance) and acceleration

  1. Oct 3, 2007 #1
    1. The problem statement, all variables and given/known data
    A baseball is thrown straight up. The drag force is proportional to v[tex]^{2}[/tex]. In terms of g, what is the ball's acceleration when its speed is half its terminal speed and a) it is moving up? b) it is moving back down?

    2. Relevant equations
    [tex]\sum F=ma[/tex]
    [tex]mg + bv^{2}=ma[/tex] - When the ball is moving up
    [tex]mg - bv^{2}=ma[/tex] - When the ball is moving back down

    3. The attempt at a solution
    We did something like this in class except we used [tex]mg - kv=ma[/tex]. We went through the calculus and got an equation for v(t) and then took the derivative to get a(t). Following my notes, I did the calculus to get an equation for v(t), and then I took the derivative to get an equation for a(t) using [tex]mg - bv^{2}=ma[/tex]. Below is what I got (in the attachment). I haven't done calculus like this since last May, so there may be some errors, but that's what I got.

    So I got an answer for a(t) for when the ball is moving down, but I don't know what to do in order to simplify it so that it's acceleration for "when its speed is half its terminal speed".

    Attached Files:

    Last edited: Oct 3, 2007
  2. jcsd
  3. Oct 3, 2007 #2
    I think you are really good at nonlinear calculus or I am missing something. How did you solve

    [tex]\int_0^t dt = \int_{v(0)}^{v(t)} \frac{dv}{g + \frac{b}{m} v^2}[/tex]
  4. Oct 3, 2007 #3
    I did my algebra before this step a little differently to get [tex]\frac{1}{m}dt=\frac{1}{mg-bv^{2}}dv[/tex]

    Then use U substitution to integrate. [tex]u=(mg-bv^{2})dv[/tex] and [tex]du=-2bvdv[/tex]

    You need to first multiply both sides by [tex]-2bv[/tex] in order to be able to use the u substitution.

    So you should have [tex]\frac{-2bv}{m}dt=\frac{-2bv}{mg-bv^{2}}dv[/tex]

    Then plug in du and u so you have

    Next integrate... [tex]\int_0^t \frac{-2bv}{m}dt = \int_{v(0)}^{v(t)} \frac{1}{u}du[/tex]

    So you get [tex]\frac{-bv^{2}t}{m}=ln\frac{mg-bv^{2}}{mg}[/tex]

    The rest is algebra to solve for v.
    Last edited: Oct 3, 2007
  5. Oct 3, 2007 #4
    Yeah, the problem is that you can't directly integrate

    [tex]\int_0^t \frac{-2bv}{m}dt [/tex]

    because there is a time dependence in v. Let me see what mathematica says.

    Supposedly the solution is

    [tex] v[t] = \frac{(\sqrt{g} \sqrt{m} Tan[\frac{(\sqrt{b} \sqrt{g} t)} {\sqrt{m}}])}{\sqrt{b}} [/tex]

    Yep, it gave the right soln. You could probably find the integral in an integral table somewhere.
    Last edited: Oct 3, 2007
  6. Oct 3, 2007 #5

    I just went off my notes (where we had [tex]mg-kv=ma[/tex]), and in my notes we did [tex]\int_0^t \frac{-k}{m}dt = \int_{v(0)}^{v(t)} \frac{-k}{u}dv[/tex], where [tex]u=mg-kv[/tex] and [tex]du=-kdv[/tex]

    We ended up getting [tex]v=\frac{mg}{k}(1-e^{\frac{-k}{m}t})[/tex]

    But this was when we had [tex]F_{air}=kv[/tex]. Now we have [tex]F_{air}=bv^{2}[/tex] and things are a bit harder.

    I now see that we have v and t in that integral so it doesn't work. Oh well, at least my paper looks all pretty with calculus all over it :).

    P.S. How do you make the LaTeX images larger? I tried the \large and \huge but it doesn't have any effect. My solution for v two posts up is difficult to read.
    Last edited: Oct 3, 2007
  7. Oct 3, 2007 #6
    I'm not really sure, but, if you haven't seen it already, there is a LaTeX for starters here


    Alternatively, just say \textrm{exp}(blah) if you don't want a cramped exponential.

    Is this a physics 1 class, classical mechanics, something else? Not a trivial differential equation to solve.
  8. Oct 4, 2007 #7
    nevermind...I got it. Was overthinking this problem too much. Just needed to solve for terminal velocity and then divide by half, then plug back into the F=ma equation.
  9. Oct 11, 2009 #8
    what was the answer to this question?
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