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Math Help - [SOLVED] vertical motion

  1. #1
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    [SOLVED] vertical motion

    The height of an object moving vertically is given by:

    s=-16t^2+96t+112

    with s in ft and t in sec. Find a) the object's velocity when t=0 b) its maximum height and when it occurs, and c) its velocity when s=0

    I don't even know where to start
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  2. #2
    Member integral's Avatar
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    Don't treat it as a physics question.

    Find the vertex ( \frac{-b}{2a})
    Then plug in two random numbers (neg and pos) to find the other two points and graph.

    (t is just x so look at the x axis for t)
    t=0 is when the parabola hits the y axis. (I.e. Just set everything multiplied by t to 0)
    max height is just the vertex.
    (and for velocity at s=0 just set the equation equal to =0)sorry I thought v=s
    Last edited by integral; January 1st 2010 at 11:50 AM. Reason: pointing out a mistake
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  3. #3
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    well i have to figure it using derivatives
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  4. #4
    Member integral's Avatar
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    use the delta process and plug in for the resulting equation 0

    velocity as t=0 is

    V=96x
    v=96t+112
    Last edited by integral; January 1st 2010 at 12:02 PM.
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  5. #5
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    so a. 96ft/s b. 256ft at t=3 and c. -128ft/sec?
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  6. #6
    Member integral's Avatar
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    so a. 96ft/s yes b. 256ft at t=3 yes and c. -128ft/sec? took me like 20 min to confirm but yes
    .
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  7. #7
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    Thank you very much for your time (twenty minutes and i might have quit if i were you) and effort. I finally get the problem... yea!!!
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