A cement block accidentally falls from rest from the ledge of a 51.5 -m-high building. When the block is 12.9 m above the ground, a man, 1.60 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?

Answer 1

V^2 = Vo^2 + 2g*d
V^2 = 0 + 19.6(51.5-12.9) = 756.56
V = 27.5 m/s.

d = Vo*t + 0.5g.t^2 = 12.9-1.6 = 11.3 m
d = 27.5*t + 4.9t^2 = 11.3
4.9t^2 + 27.5t - 11.3 = 0
Use Quadratic Formula and get:
t = 0.385 s.