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    Velocity of a falling rain drop and air resistance

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    A raindrop is observed at time t=0 when it has mass m and downward velocity u. As it falls under gravity its mass increases by condensation at a constant rate and a resisting force acts on it, proportional to is speed and equal to v when the speed is v.
    a) Show that d/dt((M^2)v) = (M^2)g where M = m+ t.
    b) Show that the speed of the raindrop at time t is (g/3 )[m+ t-((m^3)/(m+ t)^2)] + ((m^2)u)/(m+ t)^2

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    Solution Preview

    Following is the text part of the solution. Please see the attached file for complete solution. Equations, diagrams, graphs and special characters will not appear correctly here. Thank you for using Brainmass.

    Consider the situation between the times t and t + dt.

    Change in linear momentum dp = (M+ dM)(v + dv) - Mv

    dp = M dv + v dM

    (Note that dM dv term is ignored since it is very small)

    Force = dp/dt = M dv/dt + v dM/dt

    But downward force = Mg - λv


    M dv/dt + v dM/dt = Mg - λv ----------- (1)

    But, rate of increase of mass is ...

    Solution Summary

    This problem is about a falling rain drop, mass of which keeps increasing at a constant rate as it falls against air resistance. I have provided a 4-page solution to the given problem. In order to derive the first relation ship (part a) and to find the time dependence of the rain drop's velocity (part b), it requires a lot of mathematics. I have provided clear and complete solution along with a diagram to this challenging question.