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Math Help - Calculus and Baseball

  1. #1
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    Calculus and Baseball

    An outfielder fields a baseball 280 ft away from home plate and throws it directly to the cather with an initial velocity of 100 ft/s. Assume that the velocity v(t) of the ball after t seconds satisfies the differential equation dv/dt = -1/10v because of air resistance. How long does it take for the ball to reach home plate? (Ignore any vertical motion of the ball).

    I apologize if this question has already been posted. But I've spent some time searching and haven't found anything yet.
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  2. #2
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    The simple models are all the same. Somehow, cram it in your head.

    s(t) = \frac{1}{2}at^{2} + v_{0}t + h_{0}

    As long as we have only one component of acceleration, this is all we need to know. I'll ignore the units. You make sure everything works.

    Ignore any vertical motion of the ball
    h_{0} = 0

    s(t) = \frac{1}{2}at^{2} + v_{0}t

    an initial velocity of 100 ft/s
    v_{0} = 100

    s(t) = \frac{1}{2}at^{2} + 100t

    dv/dt = -1/10v
    a = -\frac{1}{10}

    s(t) = \frac{1}{2}\left(\frac{-1}{10}\right)t^{2} + 100t

    Now what?

    Hint: The solution is NOT 2000 seconds. You tell me why.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TKHunny View Post
    a = -\frac{1}{10}
    Careful!

    I have to ask. There are two possible interpretations of what the OP wrote. To the OP: Is the acceleration
    a = -\frac{v}{10} (as it is written)
    or is it
    a = -\frac{1}{10v} (as many students might write it to be)

    If it is the first case then
    a = -\frac{v}{10}

    \frac{dv}{dt} = -\frac{v}{10}

    \frac{dv}{dt} + \frac{1}{10}v = 0

    Then v(t) must be of the form
    v = Ae^{-t/10}

    Now apply your initial conditions and you are done.

    If the equation is the second of the two, that's a much more involved solution. If you need that solved, just say so and someone will help you.

    -Dan
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  4. #4
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    Calculus & Baseball - Part 2

    Thanks
    Last edited by ReneePatt; December 6th 2009 at 02:50 PM. Reason: Thought it should go in it's own thread
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  5. #5
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    Does this look correct?

    This is how I worked the problem out and what I got - is it correct?

    h'(t) = -32t + c

    v_0=h'(0)=-32(0)+c=c

    h'(t)= v_0-32t

    h(0)=0

    h(t)= v_0{t}-16t^2

    h'(t)=0

    0=h'(t)= v_0-32t

    t=\frac{v_0}{32}

    h(\frac{v_0}{32})=v_0(\frac{v_0}{32})-16(\frac{v_0}{32})^2=\frac{v_0^2}{32}-\frac{v_0^2}{64}=\frac{v_0^2}{64}

    280=\frac{v_0^2}{64}

    17920=v_0^2

    \sqrt{17920}=v_0=133.87\ sec
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  6. #6
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    Exclamation

    Quote Originally Posted by TKHunny View Post
    The simple models are all the same. Somehow, cram it in your head.

    s(t) = \frac{1}{2}at^{2} + v_{0}t + h_{0}
    Where did you get this equation?

    Quote Originally Posted by TKHunny View Post
    s(t) = \frac{1}{2}\left(\frac{-1}{10}\right)t^{2} + 100t

    Now what?

    Hint: The solution is NOT 2000 seconds. You tell me why.
    I'm not sure what to do next. How would you solve for this and NOT get 2000 seconds because that's exactly what I got?
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  7. #7
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    I'm completely confused! There seem to be two different problems being worked on here.

    For the problem I saw posted, horizontal motion with acceleration (-1/10)v, topsquarks exponential function is correct. (I did it thinking the acceleration was -1/(10v) and got a square root function.) But your last post and TKHunny's seem to be about an object dropping under acceleration due to gravity.
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    I'm completely confused!
    You and me both. I have no clue how to work this problem out and am not understanding any of the suggestions. This is due tomorrow and it looks like it'll go unanswered. I'll just have to show the different ways I tried to solve the problem and see if that gets me any where.
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