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Air resistance/velocity problem

  1. Apr 11, 2012 #1
    1. The problem statement, all variables and given/known data

    A ball is projected vertically upwards with an initial speed of 20ms^−1 at a
    height of 1.5m above the ground. Model the ball as a sphere of diameter D, mass m, and
    assume that the quadratic model of air resistance applies.

    Show that the component of acceleration at time t in the upward
    direction is given by
    a(t) = − g/b^2 x (v^2+b^2)

    where v is the speed of the ball at time t, and b2 = mg/0.2D2.

    2. Relevant equations



    3. The attempt at a solution

    There are 2 forces acting on the sphere, its weight and the air resistance, both acting vertically downwards. I used Newtons 2nd law

    ma = mg + 0.2D^2v^2

    I figure that the question involves rearranging this expression in terms of a. Now I am stuck. As I cant rearrange this expression to get the one given in the question, this leads me to believe my approach is wrong
     
    Last edited: Apr 11, 2012
  2. jcsd
  3. Apr 11, 2012 #2

    PhanthomJay

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    If you divide both sides of the equation by m to isolate a, after first substituting .2 d^2 =mg/b^2, as given, you've got it! The minus sign in the solution assumes downward as negative. You assumed downward as positive, which is fine.
     
  4. Apr 12, 2012 #3
    Thanks for the reply, but I still dont get how to do this. In particular where the extra b^2 comes from.

    I rearranged the given expression for b^2 to get -

    0.2D^2 = mg/b^2

    After substituting in

    ma = mg + (mg/b^2)v^2

    Which I get to

    a = g + (g/b^2)v^2

    Which is close, but I see no point anywhere where an extra b^2 could appear from...
     
  5. Apr 12, 2012 #4

    PhanthomJay

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    yes, now multiply the first term on the right of the equal sign by [itex]b^2/b^2[/itex], (which is 1, which doesn't change its value)
    [itex]a = g(b^2/b^2) + (g/b^2)v^2[/itex], now factor:
    [itex]a = g/b^2( v^2 + b^2)[/itex]
     
  6. Apr 12, 2012 #5
    Thanks! But how did you know to do that? Is there a rule for it?
     
  7. Apr 12, 2012 #6

    PhanthomJay

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    Gee, I don't know, I never did it this way before, I even surprised myself:eek:. If you hadn't provided the solution, I would have done it like

    [itex]a = g + (g/b^2)v^2[/itex]
    [itex]a = g(1 + (v^2/b^2))[/itex]
    [itex]a = g(1 + (v/b)^2)[/itex]

    But given the solution, I sort of worked backwards.:smile:
     
  8. Apr 12, 2012 #7
    Job done, looks like we both picked up something new! :)

    Thanks again!
     
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