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A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s(t)=96t-16t^2

a.) at what time will the ball strike the ground
b.) for what time t is the ball more than 128 feet above the ground?
c.) when will the ball reach its highest peak? how high is it above the ground?

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1 answer

  1. a)
    when is s = 0 ?
    0 = t(96 - 16 t)
    t = 0 of course, it starts at 0
    t = 96/16 = 6 seconds

    b)
    when is s = 128 (two times, on the way up and on the way down)
    128 = 96 t - 16 t^2
    8 = 6 t - t^2
    t^2 - 6 t + 8 = 0
    (t-4)(t-2) = 0
    so between 2 seconds and 4 seconds

    c)where is the vertex of this parabola?
    Well it is halfway in time between 2 seconds and four seconds, which is 3 seconds
    it is also halfway in time between 0 seconds and 6 seconds, which of course is also at 3 seconds
    so
    s(3) = 96(3) - 16(9)
    = 144 feet

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