Physics
posted by GS .
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?.

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 5439 2 =13 m.
h =vt+gt²/2
gt²+vt2H = 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