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Aircraft drops package find vertical distance

  1. Aug 22, 2008 #1
    An aircraft flying in a straight line at constant height h and speed U drops a package
    on a target at ground level. The package is subject to air resistance whose magnitude
    is kmv where v is the speed of the falling package, m its mass, and k > 0 a constant.
    (a) Show that the vertical distance that the package falls in time t is

    V(V/g(e^(-gt/V)-1)+t)

    where V is the terminal velocity of the package, and find the horizontal distance
    travelled in the same time t.
    (b) Find the cartesian equation for the path of the falling package.

    Hi i want to know what my constant k is??
    when i use k=-g/V i get the wrong answer, but when i use k=g/V i get the negative of what they want:S:S

    please help
     
  2. jcsd
  3. Aug 23, 2008 #2

    dynamicsolo

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    I think you should not start off by trying to impose an expression for k. (You can see what the dimensions should be, but we don't know what might crop up dimensionlessly...)

    Perhaps you should start with the two force equations, since this is a ballistics problem with air drag:

    [tex]ma_x = -kmv_x[/tex]
    and
    [tex]ma_y = mg - kmv_y[/tex]

    subject to the stated initial conditions
    [ v_x(0) = U , v_y(0) = 0 , y(0) = 0, and call x(0) = 0 ]
    and calling "downward" positive.
     
    Last edited: Aug 23, 2008
  4. Aug 23, 2008 #3
    Well i took upwards as positive and so my mg is -mg.
     
  5. Aug 23, 2008 #4

    dynamicsolo

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    Since you haven't shown the work you did, it is somewhat hard to judge why you aren't getting your signs to come out right. I suspect that you have not handled the differential equations and the initial conditions consistently in terms of signs.
     
  6. Aug 23, 2008 #5
    ok i will show it its just that its hard to type it all up

    Total force = -mgj -kmv
    Newtons 2nd law a = -gj -kv => a + kv = -gj
    Try integrating factor e^(int kdt)= e^(kt)


    d/dt (e^kt v) = -ge^kt j
    e^kt v = -g/k e^kt j + C

    t=0 v=u => C = u + g/k j

    v = (u + g/k j)e^-kt - g/k j

    v=dr/dt

    r = -1/k (u + g/k j) e^-kt - g/k tj + C'
    r=0 t=0 => C' = 1/k (u + g/k j)

    r = -1/k (u + g/k j) e^-kt - g/k tj + 1/k (u + g/k j)
    r = (g/k^2 j + u/k)(1 - e^-kt) - g/k tj

    u=ui as only horizotal speed

    x(t) = u/k (1 - e^-kt)
    y(t) = g/k^2 (1-e^-kt) -g/k t

    and then when i work out what k is i dont get the answer as i get k=-g/V
     
  7. Aug 23, 2008 #6

    dynamicsolo

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    Ah, but there's a problem right here. If upward is positive and the object is falling, then with the weight force being downward and the drag force upward, this needs to be

    [tex]ma_y = -mg + kmv_y[/tex]

    or just [tex]\frac{dv_y}{dt} = -g + kv_y[/tex]

    (Oh, and if you're going to write in unit vectors -- which isn't really necessary -- be sure to put them in across the board.)

    The choice of axis makes things awkward, in that you must interpret vy as negative throughout your solution. If you write the "correct" force equation, k must also be negative. If you use the equation you've written, k will be positive, but the velocity must still be treated as negative.

    It really makes for less of a headache to call downward positive and work with

    [tex]\frac{dv_y}{dt} = g - kv_y[/tex]

    but it should work either way if you treat all the signs of quantities consistently.

    BTW, an integrating factor will work, but you could just solve this as a separable differential equation.
     
  8. Aug 24, 2008 #7
    Actually what i want to know is that my way with vectors how do i split up the x and y component as with y the kmv is +ve
     
    Last edited: Aug 24, 2008
  9. Aug 24, 2008 #8

    LowlyPion

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    Drag force retards speed. Gravity is accelerating speed. Whichever convention for direction you use, your equation should reflect that they have opposite signs.
     
  10. Aug 24, 2008 #9

    LowlyPion

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    Not sure what you mean by "ve", but your x component of velocity comes from the initial speed of the plane. But of course that velocity slows as the package encounters the drag. Note that drag is proportional to velocity so as it slows the drag becomes less until it approaches zero as velocity approaches zero - in the x direction only.

    The trajectory then is the combination of the x/y position equations as a function of time.
     
  11. Aug 24, 2008 #10
    its ok thanks i get it now and +ve means positive :P sorry for the confusion
     
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