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Rain Drop VTerminal

  1. Jan 29, 2009 #1

    The terminal velocity of a 3×10−5kg raindrop is about 5m/s . Assuming a drag force , and Drag Force = -bv

    1. Assuming a drag force determine the value of the constant .

    2. Assuming a drag force determine the time required for such a drop, starting from rest, to reach 63% of terminal velocity.

    3. The attempt at a solution

    I solved Q1 with an answer of b=6*10^-5 kg/s

    for #2. I used Vfinal = Vinit + at

    Equation 1 = V final = 0.63 of Vterm and therefore is 3.15 m/s
    V init = 0
    a = ?
    Time = ?

    for a I used the constant b and did FBD and N2

    got Fdrag - mg = ma


    -(6E-5kg/s)(3.15m/s) - (3E-5kg)(9.8m/s2) = (3E-5kg)a

    but that gives a = -16.1 m/s2 which doesn't make sense to me. And i dropped that number in equation 1 above.

  2. jcsd
  3. Jan 29, 2009 #2
    Think about the direction of the forces. The raindrop is falling down, thus accelerating downwards. The drag force is slowing the drop down and is working against the direction of motion. Isn't it gravity, on the other hand, that pulls the drop downwards and therefore causes the acceleration?
  4. Jan 29, 2009 #3


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    The equation of motion for your situation is [tex] ma = -mg + F_{drag} [/tex]

    This means that you cannot use [tex] V_{final} = V_{init} + at [/tex] !!! That is the solution for motion without drag. You need to integrate the equation of motion with a drag force and use that, or the appropriate relationship derived from it, to find your answer.

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