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A basketball is shot from 2 meters up at an angle of 60° above the x axis at an initial velocity of 9 m/s.What is the maximum height the ball will reach?

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neil

  1. initial distance up = 2
    initial velocity component up = 9 sin 60 = 7.79
    v = 9 sin 60 - 9.8 t
    when v = 0, we are there
    9.8 t = 7.79
    t = .795 seconds to top
    h = 2 + 7.79(.795) - 4.9(.795^2)

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    Damon

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