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Terminal velocity/drag

  1. May 17, 2009 #1
    1. The problem statement, all variables and given/known data

    The fastest recorded skydive was by an Air Force officer who jumped from a helium balloon at an elevation of 103000 ft, three times higher than airliners fly. Because the density of air is so small at these altitudes, he reached a speed of 614 mph at an elevation of 90000 ft, then gradually slowed as the air became more dense. Assume that he fell in the spread-eagle position and that his low-altitude terminal speed is 125 mph. Use this information to determine the density of air at 90000 ft.

    2. Relevant equations

    V=sqrt(4W/pA) w= weight, p=coefficent, a = area .... D=.25pv^2A

    3. The attempt at a solution

    no idea what to do...
     
  2. jcsd
  3. May 17, 2009 #2

    LowlyPion

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    Consider the two equations at the two altitudes for terminal velocity.

    Fdrag = m*g = 1/2*Cd*p*A*v2

    Since the weight is the same ...

    1/2*Cd*p90*A*V902 = 1/2*Cd*po*A*Vo2

    p90*V902 = po*Vo2
     
  4. May 17, 2009 #3
    i got .054, however that did not work... i used 1.29 as the p for the low altitude
     
  5. May 17, 2009 #4

    LowlyPion

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    What units do they want the answer in?

    Imperial or metric?

     
  6. May 18, 2009 #5
    .054 kg/m^3.
     
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