Question

A cement block accidentally falls from rest from the ledge of a 51.7-m-high building. When the block is 18.7 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? Number Units

Answer 1

block starts from 51.7 m
its 18.7 m when he sees it
So what is the velocity at which the block is travelling when he sees it?
v= final velocity
u = initial velocity
s = distance travelled
g = gravitational acceleration
v^2 - u^2 = 2gs ( g = 9.8 m/s ; s = 51.7 - 18.7 = 33; u = 0)
v^2 - 0 = 2*9.8*33
v^2 = 646.8
v = 25.43= v1 (say)
velocity at which the block is traveling when the man sees the block is = 25.43

now the distance the block has to travel to hit his head from this point = 18.7-2 = 16.7 m (as 2 m is his height)

what is the velocity at which the block is traveling when it hits the mans head? lets cal it v2

v2^2 - v1^2 = 2 * 9.8 * 16.7
v2^2 -646.8 = 327.3
v2 = 31.2

now what is the time taken for the block to travel from 18.7m to 2m = time that the man has to get out of the way = lets say t

v = u + gt
v2 = v1 + 9.8*t
31.2= 25.43 + 9.8 *t

=> t = 0.58 secs

Therefore time the man has to get out of the way = 0.58 seconds