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A ball is thrown vertically upward, which is the positive direction. A little later it returns to its point of release. The ball is in the air for a total time of 8.26 s. What is its initial velocity? Neglect air resistance.

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    Its vertical distance is given by a parabola.
    Since we know the t-intercepts are 0 and 8.26 ,and we know a =-4.9 m/sec^2
    the equation is
    s = -4.9t(t-8.26)
    = -4.9t^2 + 40.472 , where m is in metres

    v = -9.8t + 40.472
    when t = 0 , v = 40.472

    initial velocity is 40.72 m/sec
    Wow, that is quite a toss!

    BTW, if you are working in feet,
    a = -32 ft/sec^2
    duplicate the above steps, not much has to be changed.

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