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a ball is thrown straight upward with some unknow velocity. The ball reaches a maximum height above the launch point of 32m. how fast is the ball movin when it is 8 m above the launch point

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  2. (a ball is thrown straight upward with some unknow velocity. The ball reaches a maximum height above the launch point of 32m. how fast is the ball movin when it is 8 m above the launch point )

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  3. v = Vi - g t
    0 = Vi - 9.8 t where t is time to stop at top
    so
    Vi = 9.8 t
    32 = 0 + Vi t - 4.9 t^2
    32 = 9.8 t^2 - 4.9 t^2
    t^2 = 6.53
    t = 2.56 seconds to top
    so
    Vi = 9.8*2.56 = 25 m/s

    8 = 0 + 25 t - 4.9 t^2 where ti is time at 8 m
    4.9 t^2 -25 t + 8 = 0
    t^2 - 5.11 t + 1.63 = 0
    t = [ 5.11 +/- sqrt(26.11-6.52) ]/2
    = [ 5.11 +/- 4.43]/2
    small answer is on the way up, use it
    t = .34 seconds
    v = 25 - 9.8*.34 = 21.7 m/s

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