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Terminal Velocity of a sky diver

  1. Nov 11, 2007 #1
    1. The problem statement, all variables and given/known data

    A sky diver of mass 83.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 46.0 m/s.
    (a) What is the acceleration of the sky diver when her speed is 30.0 m/s?

    2. Relevant equations
    R=-bv
    mg-bv=ma


    3. The attempt at a solution

    since R is the resistance force acting on the sky diver: R=mg
    therefore I set:
    83*9.8 = -b(46),
    and solve for b, b = -17.683
    then plug the same numbers back into mg-bv=ma to solve for the acceleration at 30m/s
    (83)(9.8)-(17.683)(30)=(83)a
    a = 3.41m/s^2


    but this is not correct, can someone please tell me what i did wrong?
     
  2. jcsd
  3. Nov 11, 2007 #2

    Dick

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    It looks fine to me. Though you usually take fluid friction proportional to v^2, not v.
     
  4. Nov 11, 2007 #3
    well, so how would I approach this question with v^2?
    I know for sure that I got the answer incorrect because I typed it into my online homework applet, it came out to be wrong.
     
  5. Nov 11, 2007 #4

    Dick

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    You'd do it exactly the same but use R=-bv^2 instead. But all of this depends on what you are supposed to assume. Did the problem ask you to take R=-bv? This is always a problem with these applets. It could be expecting a minus sign (since the acceleration is down), it could be expecting a different number of significant figures, who knows? But it you think you should take R=-bv, then you did it correctly.
     
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