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Equation of velocity with quadratic drag, vertical throw

  1. Dec 16, 2012 #1
    1. The problem statement, all variables and given/known data

    I am having trouble deriving the velocity equation for an objekt moving vertically upwards. the net force will be
    [tex]m*dv/dt=-mg-kv^2[/tex]
    where k is the drag. At the time t=0 the object's velocity wil be
    [tex]v(0)=v_0[/tex]
     
  2. jcsd
  3. Dec 16, 2012 #2

    haruspex

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  4. Dec 16, 2012 #3
    im not sure how they get to that equation, and why they are using y, which seems to be an height.
    I have tried seperating the variables, and have come to this integral
    [tex]-√(m/gk)∫1/(1+u^2 )du=∫dt[/tex]
    where u is a substitution defined by
    [tex]u=√(k/mg)⋅v[/tex]
    which leads to this
    [tex]-√(m/gk)⋅arctan(√(k/mg)⋅v)+c=t[/tex]
    however I'm not sure if this is corrrect, and the constant c must be solved for the condition t=0 and v(0)=v_0. Furthermore this will have to be solved for v, because i need the velocity equation.
     
  5. Dec 16, 2012 #4

    haruspex

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    That all looks correct. Can you not rearrange it easily enough into the form v = f(t)?
     
  6. Dec 17, 2012 #5
    for the start condition (t,v)=(0,v_0) i can only cook c down to this
    [tex]c=√(m/gk)*arctan(√(k/mg)*v_0 )[/tex]
    which is going to give a pretty nasty equation when inserted into the previous equation, that i want to solve for v?
     
  7. Dec 17, 2012 #6
    which i eventually solve to this for v
    [tex]tan((t-√(m/gk)⋅arctan(√(k/mg)⋅v_0 ))/√(m/gk))/√(k/mg)[/tex]

    or in a more pretty picture: http://imgur.com/V53iV

    but I'm not sure if it's correct, or if it can be shortened
     
  8. Dec 17, 2012 #7

    haruspex

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    arctan(v) = t+c
    v = tan(t+c) = (tan(t)+tan(c))/(1- tan(t)tan(c))
    v0 = tan(c)
    v = (tan(t)+v0)/(1- tan(t)v0)
     
  9. Dec 17, 2012 #8
    I don't really follow you there, would you mind elaborating that?
     
  10. Dec 17, 2012 #9

    haruspex

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    [tex]-√(m/gk)⋅arctan(√(k/mg)⋅v)+c=t[/tex]
    [tex]arctan(√(k/mg)⋅v)=\frac{c-t}{√(m/gk)}[/tex]
    [tex]√(k/mg)⋅v=tan(\frac{c-t}{√(m/gk)})[/tex]
    [tex]=tan(\frac{c}{√(m/gk)}-\frac{t}{√(m/gk)})[/tex]
    [tex]=\frac{tan(\frac{c}{√(m/gk)})-tan(\frac{t}{√(m/gk)})}{1+tan(\frac{c}{√(m/gk)})tan(\frac{t}{√(m/gk)})}[/tex]
    At t=0:
    [tex]√(k/mg)⋅v_0=tan(\frac{c}{√(m/gk)})[/tex]
    [tex]√(k/mg)⋅v=\frac{√(k/mg)⋅v_0-tan(\frac{t}{√(m/gk)})}{1+√(k/mg)⋅v_0 tan(\frac{t}{√(m/gk)})}[/tex]
    [tex]v=\frac{v_0-\frac{1}{√(m/gk)}tan(\frac{t}{√(m/gk)})}{1+√(k/mg)⋅v_0 tan(\frac{t}{√(m/gk)})}[/tex]
     
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