physics
posted by louis aaaa on .
A cement block accidentally falls from rest from the ledge of a 51.5 mhigh 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?

Hh= gt²/2
t=sqrt{2(Hh)/g} =
=sqrt{2(51.512.9)/9.8}=2.8 s.
v=gt =9.8•2.8=27.44 m/s
h₁=12.9 – 1.6 =11.3 m
h₁=vt₁+gt₁²/2
gt₁² + vt₁h₁= 0,
t₁² + 27.44t₁ 11.3 = 0,
t₁² + 2.8t₁ 1.15 = 0,
t₁=1.4±sqrt(1.96+1.15)
t₁=1.4±1.76
t₁=0.39 s