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Time taken for a parachute to fall

  1. May 24, 2005 #1
    Hi.

    I'm doing a KS3 coursework experiment whereby I need to investigate the question, 'Does the area of a parachute affect the rate of fall?' Obviously the answer is yes, however the theoretical physicist within me wanted to take this much further, so I attempted to pull together my basic understanding of drag, air resistance etc. in order to come up with a formula. I am using square parachute material made out of standard polyphene shopping-bags attatched to a spherical 1g weight of plasticine via. cotton. I am not taking into account the drag of the weight or cotton or the weight of the parachute material or the cotton, each of which I believed were neglieable. I came up with the formula seen in 'parachute2.JPG'. What I was trying to find, as explained better in 'Basicfor.JPG', was the length of time it takes for the parachute to reach its terminal velocity plus the time it takes for the parachute to travel the rest of the distance to the floor.

    In the formula:
    b = total time taken (seconds?)
    m = mass of weight attactched to parachute (in kg?)
    g = gravitational force (9.81?)
    C = numerical drag coefficent (1.28 for a flat plane?)
    p = air density (0.0022 approx.)
    A = crossectional area (cm sqaured?)
    '9.81' = acceleration under the affect of gravity (metres/second squared?)
    d = total distance from where parachute was dropped from to floor (cm?)
    u = initial velocity of parachute (zero m/s)
    t = time taken
    a = acceleration (9.81 m/s squared?)


    Firstly, is this formula correct? And secondly, are the units correct? I would appreciate any help in this matter.
     

    Attached Files:

  2. jcsd
  3. May 24, 2005 #2
    Please identify what 's' means.
    Terminal velocity is when the drag force exactly equals the gravitational force.
    Its best to work in MKS, meters kilograms seconds.
    Whatever units you decide, stick with them throughout.

    Also tell us how you got those equations.
     
  4. May 25, 2005 #3
    'S' (as in m/s) means 'seconds', as in m/s means metres per second.

    The equation basically works out the time taken for the parachute to reach its terminal velocity added to the time taken for the paracute to traverse the remaining distance to the floor.
     
  5. May 25, 2005 #4
    Your work makes no sense.

    In picture "BasicFor" you are subtracting 's' with units of time from 'd' with units of distance.

    I asked you to explain how you got to your conclusions but I still dont see any explaining.
     
  6. May 26, 2005 #5
    Sorry. I appear to have misunderstood your previous enquiry.

    'S' represents the distance travelled by the parachute whilst it reached mits terminal velocity.
     
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