Physics

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For beginner parachutists, the terminal velocity must be less than 5.00 m/s, and the parachutes used have a k-value that is normally distributed with a mean of 2.05/s and a standard deviation of 0.04/s. If the acceleration due to gravity is 9.81 m/s^2, determine the probability, to the nearest thousandth, that the terminal velocity is less than 5.00 m/s.

Please help me, the help in the last question was greatly appreciated. Thank-you!

  • Physics -

    Somebody please help me, I am really having problems. Thank-you!

  • Physics -

    Could you please post the relevant equations and your attempt at the solution?
    What is a k-value for a parachute?
    Thanks

  • Physics -

    Okay, relevant equations:
    v(subscript)t=g/k
    or
    k=g/v(subscript)t

    My attempt at the soloution, honestly, I have no clue, me and a friend have attempted this question and we still cannot figure it out. Please help me, I am totally lost.

  • Physics -

    so v mean = 9.81/2.05 = 4.785 m/s

    k one sigma lower = 2.05 -0.04 = 2.01/s

    velocity 1 sigma above mean = 9.81/2.01 = 4.881 /s
    so
    sigma of velocity = 4.881 - 4.785 = .095

    now 5.00 is how many sigmas above mean?
    5.00 - 4.785 = .215
    and z = .215/.095 = 2.26 sigmas above mean (pretty improbable)
    I only have a crude normal distribution table hgere but here are two values:
    z = 2.2 then F(z) = .986
    z = 2.3 then F(z) = .989
    so we are talking around .988

  • Physics -

    You and your mate would be better off reading something about terminal velocity. You cannot post questions to Jiskha in the exam hall! :)
    Here are some background reading:
    this one from NASA
    http://exploration.grc.nasa.gov/education/rocket/termvr.html
    here's some lighter reading:
    http://www.northallertoncoll.org.uk/physics/module%202/terminal%20velocity/terminal%20velocity.htm
    and here's one on statistics:
    http://en.wikipedia.org/wiki/Standard_deviation

    It's the last one that concerns you most.

    k=g/vt
    so
    vt=g/k
    g=9.81 m/s/s
    k=2.05 (mean value)
    so the mean landing velocity is
    vt=9.81/2.05=4.785
    The minimum value (kmin) of k to have a landing velocity of 5.00 m/s is
    5=9.81/km
    or
    kmin=9.81/5=1.962

    Difference from mean
    = 2.05-1.962
    =0.088

    Standard deviation (meausure of variability of the k-value)
    = 0.04

    Therefore, a parachute has to have a value of k at 0.088/0.04=2.2 standard deviations below the mean value to have a landing velocity higher than 5 m/s.

    If you look up a table of normal distribution for a tail end of 2.2 standard deviations (sigma), you will find that the probability is 0.0139, i.e. there is a 1.39% chance that the skydivers will land at more than 5 m/s.
    The table is available here:
    http://www.math.unb.ca/~knight/utility/NormTble.htm

    This is a quickie that does not really help you with your exams. Do your reading and prepare yourself accordings.

    Good luck.

  • Physics -

    Thank-you, very much!!!

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