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A Rather Simple Freefall Question

  1. Mar 22, 2007 #1
    Alright, well I haven't taken physics yet, so I set out to try and figure this problem out on myself. And the problem was, how long does it take for a person to fall 4000 feet if their terminal velocity is 120?

    Here is my process of thought to get the answer... Please correct me when I'm wrong (I know I am)

    Height = 4000 feet
    Increase rate per second = 32.15 feet
    Terminal Velocity = 120 mph = 176 feet per second

    So to determine how long it takes to get to terminal velocity...

    176 / 32.15 = 5.42 seconds to reach TV
    (Or is it 176 / 9.8 squared?)

    4000 - 176 = 3824 feet left

    3824 / 176 = 21.72 seconds to travel that distance at TV

    So 21.72 + 5.42 = 27.14 seconds to fall that distance.

    I doubt I am right, but it all seems logical and if I messed up, please correct me, thanks a ton!


    Edit: Extremely sorry for not posting it in the homework section....
     
    Last edited: Mar 22, 2007
  2. jcsd
  3. Mar 22, 2007 #2
    You're good up to here.

    Now look at your units here... 176 is ft/s. You can't subtract a velocity from a distance. You need to do 176 * 5.42, and you end up with ft., which is ~954. Then you can do 4000 - 954 = 3046 ft left.

    So just correcting that distance, you get...

    3046 / 176 = 17.3 seconds falling @ TV

    17.3 + 5.42 = 22.72 seconds total


    However... this probably wouldn't be very close to the real answer (assuming that the terminal velocity of a human IS 120 ft/s, and they stay perfectly still and rigid) because as you go faster, you experience more air resistance. As you near terminal velocity, your acceleration gradually diminishes. I'm not sure of an exact formula, air resistance isn't until next chapter!

    Hope this helps :)
     
  4. Mar 22, 2007 #3

    hage567

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    Homework Helper

    You can also look at the section of the thread https://www.physicsforums.com/showthread.php?t=110015 "Basic Equations of 1-D kinematics" for equations useful in solving this problem.

    "Increase rate per second = 32.15 feet"

    Note: this is acceleration, and it expressed in the units of ft/s/s.
     
  5. Mar 22, 2007 #4
    Excellent! Thanks man!

    But just curious, if you did factor in air resistance roughly what would it be? If you don't feel like doing it, that's fine, the help you provided was adequate.
     
  6. Mar 23, 2007 #5
    I wish I could tell you, but I haven't learned how to yet. I won't know how to for another week or so. Air resistance starts next chapter. In order to calculate air resistance, we'd need a lot more information (drag coefficient, cross-sectional area, etc). My guess is that it would probably take about 2-3 seconds longer than our theoretical value, but that's just a guess.
     
  7. Mar 23, 2007 #6
    K, but thanks anyways. I guess I'll just have to wait until I take Physics next year! ^_^
     
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