Asked 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?.
Answers
Answered by
Elena
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
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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