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Verticle projectile motion and terminal velocities

  1. Sep 7, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle is projected vertically upward in a constant gravitational field with an initial speed [tex] v_0[/tex]. Show that if there is a retarding force proportional to the square of the instantaneous speed, the speed of the particle when it returns to the initial position is [tex] \frac{v_0 v_t}{\sqrt{v_0^2 + v_t^2}} [/tex] where [tex] v_t [/tex] is the terminal speed.


    2. Relevant equations
    Newton's 2nd law: [tex] F_{net} = m \ddot{z} [/tex]



    3. The attempt at a solution
    Since our retarding force is proportional to the square of the velocity, [tex] F_{r} = kmv^2 [/tex]
    Now putting this and gravity into newton's 2nd law, we have:
    [tex] -mg - kmv^2 = m \ddot{z} [/tex]
    therefore, [tex] \ddot{z} = -g-kv^2 [/tex]
    Separating variables and noticing that [tex] \ddot{z} = v [/tex], we have:
    [tex] \frac{dv}{g+kv^2} = -dt [/tex]
    Integrating,
    [tex] \frac{tan^{-1}(\frac{\sqrt{k}}{\sqrt{g}}v)}{\sqrt{g}\sqrt{k}} = -t + c_{1} [/tex]
    At [tex] t = 0 [/tex], [tex] \dot{z} = v = v_0 [/tex]
    Therefore, solving for [tex] c_1 [/tex] and plugging back in, we get:
    [tex] \frac{tan^{-1}(\frac{\sqrt{k}}{\sqrt{g}}v)}{\sqrt{g}\sqrt{k}} = -t + \frac{tan^{-1}(\frac{\sqrt{k}}{\sqrt{g}}v_0)}{\sqrt{g}\sqrt{k}} [/tex]

    Solving for [tex] v [/tex], I get:
    [tex] v = \frac{\sqrt{g} tan({tan^{-1}[\frac{\sqrt{k}}{\sqrt{g}}v_0] - t \sqrt{g} \sqrt{k}) }}{\sqrt{k}} [/tex]

    Integrating to find the displacement equation , we get:
    [tex] z = \frac{ln(cos(\sqrt{g}\sqrt{k} t - tan^{-1}(\frac{\sqrt{k}}{\sqrt{g}}v_0)))}{k} + c_2 [/tex]

    When [tex] t = 0 [/tex], [tex] z = 0 [/tex], so solving for [tex] c_2 [/tex] and plugging back in, we get,

    [tex] z = \frac{ln(cos(\sqrt{g}\sqrt{k} t - tan^{-1}(\frac{\sqrt{k}}{\sqrt{g}}v_0)))}{k} - \frac{ln(\frac{g}{g+kv_0^2})}{2k} [/tex]

    So, now I set this equal to [tex] 0 [/tex] so that i can find the time when the projectile returns to its initial position of [tex] z = 0 [/tex].
    Solving for [tex] t [/tex] when [tex] z = 0 [/tex], I get some extremely nasty stuff according to Mathematica.

    The equations start getting so nasty, I can't really do anything with them. Much less take the limit of v as t -> infinity to try to find the terminal velocity.



    There has to be an easier way to do this, as my professor never even mentioned using Mathematica or computers in general.

    Any help with attacking this problem would be appreciated extremely. (I can give the rest of the silly equations I got at the end if anyone wants, but it got so long that it wasn't helpful at all i don't think).
     
  2. jcsd
  3. Sep 8, 2009 #2

    kuruman

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    You need to introduce the terminal velocity in your differential equation before you solve it. This will make the algebra simpler. You can find the terminal velocity without having to let t go to infinity in some expression. Hint: What happens to the acceleration when the particle reaches terminal velocity? What happens to your diff. eq. when that happens to the acceleration?
     
    Last edited: Sep 8, 2009
  4. Sep 8, 2009 #3
    I actually realized a couple things last night while I was trying to sleep.

    First off, since my retarding force is proportional to the square of the velocity, just making it negative (like I had been doing for problems where the retarding force was proportional to the velocity itself) will not work.

    We always want the retarding force to be opposite the motion, so it seems I may have to do this problem in two steps.

    The projectile's trip all the way up would be given by [tex] -mg -mkv^2 = m \ddot{z} [/tex]. This would be used until the projectile stops at the top of its trip (velocity becomes zero). Then on the trip down, the equation would be [tex] -mg + mkv^2 = m \ddot{z} [/tex] since the retarding force needs to be up in this case.


    Now, going back to what you said.
    Terminal velocity happens when there is no acceleration, correct? So I need my two forces to be equal.

    [tex] mg = mkv^2 [/tex]
    which gives me
    [tex] g = kv^2 [/tex]
    [tex] v^2 = \frac{g}{k} [/tex]
    [tex] v = \pm \sqrt{\frac{g}{k}} [/tex]

    I don't know how that is overly helpful though, since I don't think I can solve the equations with a zero acceleration (because at some points during the projectile's trip there is definitely an acceleration)
     
  5. Sep 8, 2009 #4

    kuruman

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    You are correct in saying that you need to solve the problem in two parts. The diff. eq. for the trip up is not the same as the diff. eq. for the trip down. Introducing the terminal velocity gives the following equation for the trip up:

    [tex] \dot{v_{z}}= -g\left(1+\frac{v^{2}_{z}}{v^{2}_{T}} \right) [/tex]

    Then you use the transformation (prove it if you have not already done so)

    [tex] \dot{v_{z}}= v_{z}\frac{d v_{z}}{dz} [/tex]

    to get a differential equation without time as a variable. Solve it to get the height as a function of velocity. From this you can get the maximum height in terms of the initial velocity, the terminal velocity and g. Once you have that, repeat for the trip down and finish the question.
     
  6. Sep 8, 2009 #5
    How did you get the first equation? I can't get that.
     
  7. Sep 8, 2009 #6

    kuruman

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    Straightforward application of the chain rule of differentiation

    [tex] \dot{v}_{z}=\frac{dv_{z}}{dt}=\frac{dv_{z}}{dz}\frac{dz}{dt}=v_{z}\frac{dv_{z}}{dz}[/tex]
    :wink:
     
  8. Sep 8, 2009 #7
    Oh no, I got that one. I meant the first equation ( I am having trouble getting the terminal velocity into the differential equation). I also notice that you lost time somewhere :)

    I got the 2nd equation though. Chain rule is cool :)
     
  9. Sep 8, 2009 #8

    kuruman

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    [tex]\ddot{z}=-g-kv^{2}[/tex]

    [tex]\ddot{z}=-\left(g+kv^{2} \right)[/tex]

    [tex]\ddot{z}=-g\left(1+\frac{k}{g}v^{2} \right)[/tex]

    But

    [tex]\frac{k}{g}=\frac{1}{v^{2}_{T}}[/tex]

    Therefore

    [tex]\ddot{z}=-g\left(1+\frac{v^{2}}{v^{2}_{T}} \right)[/tex]

    QED
     
    Last edited: Sep 8, 2009
  10. Sep 8, 2009 #9
    neat. I think that's everything I'll need unless I run into another problem.
    Thanks a lot.

    Edit: Solved it!.. Wooooo!
     
    Last edited: Sep 8, 2009
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