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Another Grade 11 Dynamics Problem Needing Urgent Help!

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data

    The drag coefficient works like the coefficient of friction when an object is moving through the air, but it increases its effects as the object moves faster. A skydiver plus the parachute has a combined mass of 94.0kg. if the formula for the drag coefficient of the parachute is F_fric=C_D v^2, find the terminal velocity of the 54.0kg/m



    2. Relevant equations

    m=94.0kg a=9.8m/s^2
    Cd=54.0kg




    3. The attempt at a solution

    I do not understand the question and am not even sure what terminal velocity is in the first place
     
  2. jcsd
  3. Nov 15, 2009 #2

    ideasrule

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    As an object falls under gravity, air resistance gets stronger and stronger. Eventually, at a certain speed, air resistance will equal the force of gravity and the object stops accelerating. So what's the first step?
     
  4. Nov 15, 2009 #3
    Oh ok how do you find out when that point is?
     
  5. Nov 15, 2009 #4

    ideasrule

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    At that point, the force of gravity is equal to the force of air resistance. Write that out as a formula and you're basically done.
     
  6. Nov 15, 2009 #5
    Fg=Fa?
     
  7. Nov 15, 2009 #6

    ideasrule

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  8. Nov 15, 2009 #7
    so there isnt any answer?
     
  9. Nov 15, 2009 #8

    ideasrule

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    Fg=?
    Fa=?

    Write them out using the information you know and the information you want to find (namely, Cd, v, m, and g).
     
  10. Nov 15, 2009 #9
    ok ill try and post back
     
  11. Nov 15, 2009 #10

    ideasrule

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    You already got one:

    F_fric=C_D v^2

    What's Fg?
     
  12. Nov 15, 2009 #11
    Fg is 9.8m/s/s
     
  13. Nov 15, 2009 #12

    ideasrule

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    That's g, the acceleration due to gravity. F=ma, so Fg=?
     
  14. Nov 15, 2009 #13
    921.2m/s/s?
     
  15. Nov 16, 2009 #14
    wow i need serious help with this
     
  16. Nov 16, 2009 #15

    ideasrule

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    Fg=mg. Remember this.
     
  17. Nov 16, 2009 #16
    ok i think i might have gotten it....is this all i have to do:
    m=94.0kg
    a=9.8m/s^2
    CD=54kg/m
    v=?
    F_fric=?
    Step 1
    F_fric=ma
    F_fric=94.0*9.8
    F_fric=921.2m/s
    Step 2:
    921.2=54*v^2
    921.2/54=v^2
    17.1=v^2
    √17.1=v
    4.1=v
    v=4.1m/s
     
  18. Nov 16, 2009 #17
    is it right?
     
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