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Math Help - Upward vertical movement with air drag.

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    Upward vertical movement with air drag.

    \hat{} Hi, here's the problem: "upward vertical movement with air drag"
    speed = v; initial speed at t0 = v0; initial height x0 = 0
    x= vertical distance travelled

    force gravity Fg=-m.g
    force drag Fd=-k.v^2 (k=ct describing air, shape of object,...)

    F=m.a :
    --> m.a=-m.g-k.v^2 (a=x'' and y=x' and devide by mass m)
    --> x'' = -g - (k/m).x'^2
    --> x'' + (k/m).x'^2 = -g
    --> what is the function of x in time, and of v in time?
    That's a DE with second order and second degree, never learned this but am eager to learn it, searched som sites about DE but didn't easily find a way to solve this...
    thanks for the help!
    Bart
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  2. #2
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    Quote Originally Posted by bartroels View Post
    \hat{} Hi, here's the problem: "upward vertical movement with air drag"
    speed = v; initial speed at t0 = v0; initial height x0 = 0
    x= vertical distance travelled

    force gravity Fg=-m.g
    force drag Fd=-k.v^2 (k=ct describing air, shape of object,...)

    F=m.a :
    --> m.a=-m.g-k.v^2 (a=x'' and y=x' and devide by mass m)
    --> x'' = -g - (k/m).x'^2
    --> x'' + (k/m).x'^2 = -g
    --> what is the function of x in time, and of v in time?
    That's a DE with second order and second degree, never learned this but am eager to learn it, searched som sites about DE but didn't easily find a way to solve this...
    thanks for the help!
    Bart
    Hint:

    It may be easier to solve this problem in two parts

    Let the acceleration be

    a=\frac{dv}{dt}

    Now you can write this as a first order ode

    m\frac{dv}{dt}=-mg-kv^2

    This equation is first order and separable!

    Then use

    v=\frac{dx}{dt}

    to finish!
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