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a man is 1.8m tall and is standing at the bottom of a 54m tall building, he looks up and sees a brick that has fallen 39m towards him. how much time does he have to get out of the way before the brick hits him?.

  • Physics -

    After covering 39 m the velocity of the brick was
    v=sqrt(2gH) = sqrt(2•9.8•39) =27.65 m/s.
    The brick has to cover the distance 54-39 -2 =13 m.
    h =vt+gt²/2
    gt²+vt-2H = 0
    9.8• t²+27.65•t -26 = 0
    t =-27.65 ±sqrt(27.65²+4•9.8•26)/2•9.8 =
    = - 27.65±42.23/19.6
    t=0.74 s

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