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    1) If an arrow is shot straight upward with a velocity of 96 feet per second (ft/sec) from an altitude of 6 feet, for how many seconds is this arrow more than 86 feet high?

    If an object is given an initial velocity straight upward of "v lower case 0" feet per second from a height of "s lower case 0" feet, then its altitude "S" after "t" seconds is given by the formula:

    S = -16t squared + v lower case 0 t + s lower case 0

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    Solution Preview

    S = -16t ^2 + Vo t + So

    So= 6 feet
    Vo= 96 ft/sec

    we have to check when S>86
    or S = -16t^2+96 t +6 >86
    or -16t^2+96 t -80 ...

    Solution Summary

    The solution determines the time for which an arrow shot straight upward is above a particular height.