A cement block accidentally falls from rest from the ledge of a 53.4-m-high building. When the block is 19.4 m above the ground, a man, 2.00 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

The time required by the man to get out of the way is 0.6 s.

Explanation:

height of building, H = 53.4 m

height of block, h = 19.4 m

height of man, h' = 2 m

Let the velocity of the block at 19.4 m is v.  

use third equation of motion

Now let the time is t.

Use second equation of motion

Time cannot be negative so time t = 0.6 s.

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