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Model for air resistance

  1. Dec 31, 2005 #1
    My maths teacher before we broke up for christmas asked the class to work out a simple model for air resistance for an object that is dropped given that the resistive force is proportional to the square of the velocity. I've spent about 10 minutes thinking about it, and this is what I came up with. (I hav'ent read anything on the matter, so any suggestions, corrections, improvements etc would be kind)

    Since without air resistance, 0.5mv^2 = mg(h0 - h1)
    (h0 > h1)
    therefore with air resistance that is proportional to the square of the velocity, that means that 0.5mv^2 + mg(h1 - h0) + k(h0 - h1)v^2 = 0
    Therefore, v^2 = (mg(h0 - h1)) / (0.5m + k(h0 - h1))

    I tried this when m = 5 kg, h0 - h1 = 1 metre, k = 1.5, so if no air resistance, v = sqrt(19.6), but if air resistance is taken into account, then v = 3.5 m/s.
    This makes sense, but ideas would be helpful.
  2. jcsd
  3. Dec 31, 2005 #2


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    Air resistance is proportional to the square of the velocity, or R = kv². That all. There is no h0 or h1 involved.

    The air resistance is a force which reduces the accelerating force on a moving body.

    If a body is falling through the air, then the forces acting on it are 1: its weight = mg, acting downwards, and 2: the air resistiance R = kv², acting up, opposite to the direction of motion.

    So, the effective accelerating force is: F = mg - R.
    From newton's 2nd law,

    F = ma
    mg - kv² = ma
    mg - kv² = mdv/dt

    Edit: corrected the typo. Thanks gamma.
    Last edited: Dec 31, 2005
  4. Dec 31, 2005 #3

    I see why you wrote the above. You tried to use the conservation of energy. However, the problem is the air resistance is propotional to the velocity and velocity is not constant. Rather it increases with ((h1 -h0). So your equation is incorrect. Right way is to do what Fermat has suggested. You have to solve the differential equation.

    There is a typo in Fermat's posting. DE should be

    mg - kv² = m dv/dt


  5. Dec 31, 2005 #4


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    To phrase what Gamma just said in a different way, air resistance is not a "conservative force" and so total kinetic and potential energy is not conserved. Better to set mass times acceleration equal to total force- the equation Gamma gave. By the way, for small, relatively light, objects, air resistance is proportional to the square of speed. For heavier objects it is proportional to speed itself.
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