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Jumping Into Fire Net

  1. Jun 26, 2003 #1
    A person jumps from a fourth story window 15 m above a safety net. THe jumper stretches the net 1.0 m before coming to rest. What was the deceleration experienced by the jumper?

    Equation:

    x = x0 + v0t +.5at^2

    15 m = 1 m + 8.6 m/s(2s) + .5(a)(1.75s^2)

    15m = 1 m + 17.2 m/s^2 + .875s^2(a)

    -3.2 m = .875 s^2

    -3.66 m/s^2 = Deceleration

    Does this look right? I'm not sure about the 1 m as the final position...
     
  2. jcsd
  3. Jun 26, 2003 #2
    Wait, I already see a problem, forgot to square the time...

    Equation:

    x = x0 + v0t +.5at^2

    15 m = 1 m + 8.6 m/s(2s) + .5(a)(1.75s^2)

    15m = 1 m + 17.2 m/s^2 + 1.53s^2(a)

    -3.2 m = 1.53 s^2

    2.09 m/s^2 = Deceleration

    Does this look right? I'm not sure about the 1 m as the final position...That seems like an awfully slow deceleration for jumping out of a window.
     
  4. Jun 26, 2003 #3

    enigma

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    You've got a few errors, and I'm not sure where you got all of your values.

    Where did the 8.6m/s , the 2s, and the 1.75s^2 come from?

    You're also making mistakes with the units.

    '8.6 m/s * s' does not have units of 'm/s^2', it is 'm'

    Your tipoff that you are making mistakes comes in the line where you have -3.2m = .875 s^2

    those units don't equal each other, therefore you know you must have made a mistake somewhere.

    To do this problem correctly, you need to break it into 2 parts.

    First part is the freefall. You know starting and ending positions (15m and 0m above the net), you know starting velocity (0m/s), and you know acceleration (-9.8m/s^2). You use those numbers to solve for the velocity before the guy hits the net.

    The second part is the deceleration. You have starting and ending positions (0m and -1m), you know starting velocity (from part I), and you know final velocity (0m/s... he is stopped at the bottom). You use those values to solve for acceleration.
     
  5. Jun 26, 2003 #4
    Revised: (I always have trouble seeing problems in two distinct parts)

    v^2 = v^20 + 2a(x-x0)
    v^2 = 0 m/s + 2(9.80 m/s^s)(15m -0m)
    v^2 = 294 m^2/s^s
    Squareroot(v^2) = Squareroot(294 m^2/s^2)
    v = 17.14 m/s

    v^2 = v^20 + 2a(x-x0)
    0 m^2/s^2 = 17.14 m/s + 2a(-1 - 0)
    0 m^2/s^2 = 294 m^2/s^s -2a
    -294 m^2/s^2 = -2a
    a = 147 m/s^2

    Thats more than 14 gs!!! Can that be right? Wouldnt the person be dead?
     
  6. Jun 26, 2003 #5

    enigma

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    15 g's.

    The net is slowing him down to zero velocity in 1/15th the distance.

    Not a very good net - He would probably be hurt, but not neccessarily killed.

    15 g's is enough to cause damage, but it won't last for a very long period of time. Aircraft ejector seats pull over 100 g's for a split second.

    Is the result reasonable? If you don't have a clear mental picture of meters, convert to feet. This guy is falling from a fourth (probably closer to 5th) story window, and stopping in 3 feet. That's a pretty stiff net.
     
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