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Projectile Subject to Quadratic Air Resistance

  1. Sep 2, 2014 #1
    1. The problem statement, all variables and given/known data
    A projectile that is subject to quadratic air resistance is thrown vertically up with initial speed v0.

    (a): Write down the equation of motion for the upward motion and solve it to give v as a function of t.
    (b): Show that the time to reach th top of the trajectory is ttop = (vter/g)arctan(v0/vter.

    2. Relevant equations

    ma = -cv2

    3. The attempt at a solution

    I think I may have done something wrong with part a which is leading me astray for part b.

    Part (a):

    m dv/dt = mg - cv2
    vter occurs when mg = cv2
    vter = sqrt (mg / c)
    solving for c, I get: c = (mg / (vter)2)
    Putting this back in to my differential equation and cancelling the mass out:
    dv/dt = g(1 - (v2 / vter2))
    Integrating both sides, I get:
    gt = vterarctanh(v/vter) - vterarctanh(v0/vter)

    solving for v as a function of t, I get:

    v(t) = vtertanh((gt/vter) + arctanh(v0/vter))

    Therefore, for part b, when it is at the top of its flight, v(t) = 0. Substituting in the zero and solving for t, I get:

    (vter/g)arctanh(v0/vter)

    As you can see, this is almost what the correct answer is, except it should be a function of the inverse tangent of v0 / vter, NOT the inverse hyperbolictangent.

    Any ideas where I went wrong or what I am missing here?

    Thanks in advance.
     
  2. jcsd
  3. Sep 2, 2014 #2

    haruspex

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    Be careful with the signs. Strictly speaking, the quadratic drag equation (omitting gravity for now) is m dv/dt = -kv.|v|. This arranges that the drag always opposes the motion.
    Alternatively, to eliminate the modulus sign, use different equations for the upward and downward phases.
    If we take up as positive and g as therefore having a negative value, the upward equation is as you wrote above, but you'll need the downward equation to get terminal velocity.
     
  4. Sep 2, 2014 #3
    Ok, that makes sense, but I am a little confused as to how to create another differential equation for the downward equation. Would I just flip the sign and make it -mg-cv 2 ? The only thing I can think of is to make g negative and keep cv 2 negative since drag always opposes motion. Is that the right idea?
     
  5. Sep 2, 2014 #4
    For solving (a) and (b), you just need the equation for upward motion, which is:
    [tex]
    m\frac{dv}{dt} = -mg - cv^2
    [/tex]
    if the upwards direction is considered positive.

    Your procedure from that point on is fine until you try to solve the first order ODE. It would be very helpful (to me at least) if you could show, in detail, how you go about it.
     
  6. Sep 2, 2014 #5

    haruspex

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    Except that (b) involves "vter", for which you need to consider the downward motion.
    It depends how you define 'g'. My preference is that it is the acceleration due to gravity as a signed quantity, so the upward equation is ##m\frac{dv}{dt} = mg - cv^2##, and the downward is ##m\frac{dv}{dt} = mg + cv^2##, where g has a negative value.
     
  7. Sep 2, 2014 #6
    I might have went over the finer points a bit too quickly here. I considered 'g' a positive constant, since that was consistent with the solution given, but I shouldn't impose that on the OP.

    I also took vter to mean the magnitude of the terminal velocity.
     
  8. Sep 2, 2014 #7
    Got it! Thanks for all of the help! It ended up working out when I fixed the sign issue so that the differential equation represents the upward motion rather than the downward motion. Thank you for all of your help!
     
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