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Terminal speed-Skydiver

  1. Oct 14, 2015 #1
    1. The problem statement, all variables and given/known data
    I attached a screen shot of the problem. Screen Shot 2015-10-14 at 9.40.16 PM.png

    2. Relevant equations
    D= 1/2 C p A v^2
    bv=mg

    3. The attempt at a solution
    I spent 3 long hours on this problem. I confused myself even more. Please help me understand what is going on.
     
  2. jcsd
  3. Oct 15, 2015 #2

    Simon Bridge

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    Please show your best attempt with reasoning.
     
  4. Oct 15, 2015 #3
    I know that at terminal velocity, bv=mg ... so, mg=1/2CpAv^2.. would I just solve for v? .. v=sqrt((2mg)/(CpA))... v=1.38m/s ?
    m: 84kg
    C: 0.8
    p: 1.2kgm^3
    A: 987cm^2
     
  5. Oct 15, 2015 #4

    haruspex

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    That's the right approach, but I don't understand how you get such a small velocity. Did you forget to convert the area to m2?
     
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