I am a little confused on how to solve this, could someone help me out? A cement block accidentally falls from rest from the ledge of a 73.2-m-high building. When the block is 17.0 m above the ground, a man, 1.70 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?

Question

anonymous · Answer

Here initial vel, u=0  as it falls from rest
accl, a=-g=-9.8
So 
$$S=u*t+0.5*a*t^2$$
becomes
$$S=-0.5*9.8*t^2$$

anonymous · Answer

Time to reach 17m, t1:
Here displacement, S=17-73.2
t=t1
Substitute in the above eqn and find t1...

anonymous · Answer

Time to reach 1.7m, t2:
S=1.7-73.2 ; t=t2
find t2....

anonymous · Answer

Time of travel from 17m to 1.7 m= t2-t1
He must evade before this time to save himself...