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Equation for upward motion of the particle with air drag

  1. Jul 15, 2014 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m is thrown upward with velocity v and there is retarding air resistance proportional to the square of the velocity with proportionality constant k. if the particle attains a maximum height after time t, and g is the gravitational acceleration, what is the velocity v?

    2. Relevant equations



    3. The attempt at a solution
    if i take D=kv2 and now two forces act on the particle F=mg and the retarding force but the options here are a) √(k/g)tan[√(g/k)t]
    b)√(gk) tan [√(g/k)t]
    c) √(g/k) tan[√(gk)t]
    d) √(gk) tan[√(gk)t]
    so i m having problem in figuring out how this tan factor come in .......
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 15, 2014 #2

    BiGyElLoWhAt

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    Let me just clean this up:

    If I take ##D = kv^2 ## and now two forces act on the particle ##F = mg## and the retarding force (##D##?) but the options here are:
    a) ##\sqrt{k/g}\ tan(\sqrt{g/k}\ t)##
    b) ##\sqrt{gk}\ tan(\sqrt{g/k}\ t)##
    c) ##\sqrt{g/k}\ tan(\sqrt{gk}\ t)##
    a) ##\sqrt{gk}\ tan(\sqrt{gk}\ t)##

    right?
    Is the [edit]retarding[edit] force you're referring to ##D##?

    Where did you get these options? Are these supposed to be final answers? Multiple choice question I'm assuming, no?
     
    Last edited: Jul 15, 2014
  4. Jul 15, 2014 #3

    haruspex

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    So what equation do you get for the motion?
     
  5. Jul 15, 2014 #4

    AlephZero

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    Did you make a typo in the answer options, or have you defined ##k## wrongly?

    I don't think any of the options have the correct dimensions. For example the expressions inside ##\tan(\cdots)## are not dimensionless.
     
  6. Jul 15, 2014 #5

    haruspex

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    How true... looks like the question should have said the drag force is mkv2; or maybe k in each option should be replaced by k/m.
    Btw, if we assume that one of the options is then correct, the question can be answered merely by selecting the one with the right dimensions everywhere. There is no need to find or solve any equations of motion. Quick, but somewhat unsatisfactory, since we never get to discover how the tan function comes into it.
     
  7. Jul 15, 2014 #6

    AlephZero

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    That doesn't make much physical sense, because the drag force depends on the shape of the object but not on its mass.

    Maybe the best option is wait for the OP to come back.
     
  8. Jul 15, 2014 #7

    haruspex

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    I did not mean to suggest that such a k would have a physical meaning. I agree that the choice of k as the letter suggests the force should be kv2, because k is commonly used that way in this context. But given that m is fixed here, there's nothing to stop the question setter from choosing to define k as force/(mv2).
     
  9. Jul 17, 2014 #8
    this question is from JEST 2013 (joint entrance screening test) and the answer is supposed to be (c) according to the answer key.
     
  10. Jul 17, 2014 #9

    haruspex

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    If we ignore the 'missing mass' problem, and take the drag equation to be F = mkv2, option (c) is the only one that's dimensionally correct. Do you see that?
     
  11. Jul 18, 2014 #10
    oh!.. yes never thought of taking F = mkv2..... now i see it....thanks
     
  12. Feb 16, 2016 #11
    My guess is
    Fnet = -Fweight - Fdrag = m ⋅ dv/dt
    -m ⋅ g - k ⋅ v² = m ⋅ dv/dt
    dv/(k / m ⋅ v² + g) = -dt

    Notice that
    ∫1/(1 + x²)dx = atan(x)
    Then
    v(t) = √( g ⋅ m / k ) ⋅ tan( -t ⋅ √( g ⋅ k / m ))
    assuming initial velocity is 0
     
  13. Feb 16, 2016 #12

    haruspex

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    Yes, but the difficulty was that does not match any of the offered choices. To get one of those, you have to realise that k here need not be what you expect. You are only told it is a constant of proportionality, so it need not conform to the equation "drag=kv2".
     
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