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I need a little help attacking a problem

  1. Jun 7, 2005 #1
    hello,

    what would be the best way to approach this problem

    A skydiver is jumping out of a plane at :

    Initial velocity = 100 mph

    Initial position = 2000 m

    acceleration due to gravity = -9.81

    I need help finding air friction at different positions, change in acceleration and change in velocity ?

    i think u calculate air friction
    A= f * v * v(abs) * cd

    A= acceleration

    f= atmospheric pressure

    v= velocity

    s= surface area

    cd= drag coefficient
     
  2. jcsd
  3. Jun 7, 2005 #2

    Pyrrhus

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    Start off with Newton's 2nd Law (For constant mass) in scalar form:

    [tex] \sum_{i=1}^{n} F_{i} = ma [/tex]

    rewrite it as

    [tex] \sum_{i=1}^{n} F_{i} = m \frac{dv}{dt} [/tex]

    or

    [tex] \sum_{i=1}^{n} F_{i} = m \frac{d^{2} x}{dt^{2}} [/tex]

    And solve the ODEs formed
     
  4. Jun 8, 2005 #3
    the general form of that equation looks familiar, but isn't it actually:

    [tex]F(v) = c \rho A v^2[/tex]

    where [itex]A[/tex] is the surface area? Acceleration would then be:

    [tex]a(v) = \frac{c \rho A v^2}{m} = \frac{c \rho A}{m} ({\frac{dx}{dt}})^2[/tex]

    integrating with respect to time will give the velocity as a function of time, but this is a 2nd order DE so you need to solve the DE and plug in your initial value conditions.
     
  5. Jun 8, 2005 #4

    dextercioby

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    Not really II-nd order.Use Newton's second law in vector and the definition of linear momentum.

    Daniel.
     
  6. Jun 8, 2005 #5
    Normalize your units first off.
     
  7. Jun 8, 2005 #6

    dextercioby

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    There's no atmospheric pressure in the drag force,just the air's density...

    Daniel.
     
  8. Jun 8, 2005 #7

    arildno

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    Dearly Missed

    Nope, this is false; air resistance always works in the opposite direction of the velocity; that's why you need the absolute value sign here.
     
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