1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I need a little help attacking a problem

  1. Jun 7, 2005 #1

    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


    User Avatar
    Homework Helper

    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]


    [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


    User Avatar
    Science Advisor
    Homework Helper

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

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


    User Avatar
    Science Advisor
    Homework Helper

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

  8. Jun 8, 2005 #7


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook