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Math Help - falling body

  1. #1
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    falling body

    A free falling body has no initial velocity. Its mass is 80kg and the air resistence is proportional to the velocity squared with the proportional constant 0.16kg/m.Let velocity be y m/s and x s. And the question is to put up a differential equation for this.

    My attempt at the solution.
    y(0)=0 (easy part). Now my thinking about the equation itself goes something like this.
    F=ma
    a=9.82
    F(t) = F-Fr(t) (Fr=Air resistance)
    ma(t)=ma- v^2 * 0.16(t)
    divide it all by m which is 80
    a(t) = 9.82 - 0.002v^2(t)
    swapping out v(t) for y and a(t) for dy/dt
    dy/dt=9.82 - 0.002y^2
    which is correct according to the text book.
    My issue is That it seems that i have solved for the accelration and not for the velocity itself. Im thankful if anyone could explain this.
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  2. #2
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    Quote Originally Posted by Zamzen View Post
    A free falling body has no initial velocity. Its mass is 80kg and the air resistence is proportional to the velocity squared with the proportional constant 0.16kg/m.Let velocity be y m/s and x s. And the question is to put up a differential equation for this.

    My attempt at the solution.
    y(0)=0 (easy part). Now my thinking about the equation itself goes something like this.
    F=ma
    a=9.82
    F(t) = F-Fr(t) (Fr=Air resistance)
    ma(t)=ma- v^2 * 0.16(t)
    divide it all by m which is 80
    a(t) = 9.82 - 0.002v^2(t)
    swapping out v(t) for y and a(t) for dy/dt
    dy/dt=9.82 - 0.002y^2
    which is correct according to the text book.
    My issue is That it seems that i have solved for the accelration and not for the velocity itself. Im thankful if anyone could explain this.
    Using the symbol v instead of the (stupid) symbol y:

    dv/dt = 9.82 - 0.002v^2

    which, perhaps more obviously now, is obviously a DE for the velocity. Solve it and you get v.
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