Question

A cement block accidentally falls from rest from the ledge of a 75.1-m-high building. When the block is 17.9 m above the ground, a man, 1.80 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

U = initial velocity = 0 m/sec

a = g

H = 75.1 m

Using equation

H = U*t + 0.5*a*t^2

t = sqrt (2H/g)

t = sqrt (2*75.1/9.81) = 3.91 sec (it will reach ground)

when block is 17.9 m above ground = 75.1 - 17.9 = 57.2 m from top, at that point it's speed will be

V^2 = U^2 + 2*a*S

V = sqrt (2*9.81*57.2) = 33.5 m/sec

Now from this point initial speed = 33.5 m/sec

a = g

d = 17.9 - 1.80 = 16.1 m

time taken in travelling this distance will be

d = u*t + 0.5*a*t^2

16.1 = 33.5*t + 0.5*9.81*t^2

4.905*t^2 + 33.5t - 16.1 = 0

Solving above quadratic equation

t = [-33.5 +/- sqrt (33.5^2 + 4*4.905*16.1)]/(2*4.905)

take positive sign

t = 0.451 sec

Man will have 0.451 sec to get out of the way