Question

A cement block accidentally falls from rest from the ledge of a 73.9-m-high building. When the block is 12.4 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

To determine the time the man has, let’s find out the block’s velocity at the 12.4 meters above the ground and 1.8 meters above the ground.

Then determine the time for the block’s velocity to increase from the 1st velocity to the 2nd velocity.

Use the following equation to determine the block’s velocity at 12.4 m above the ground.
vf² = vi² + 2*a*d,

vi = 0, a = 9.8 m/s², d = 73.9 – 12.4

= 61.5 m
vf² = 2 * 9.8 * 61.5
vf = √1205.4
The velocity is approximately 34.72 m/s.

Let’s use the same equation to the block’s velocity at 1.8 m above the ground.
d = 73.9 – 1.8

= 72.1 m
vf² = 2*9.8*72.1
vf = √1413.16
The velocity is approximately 37.6 m/s.

Use the following equation to determine the time.

vf = vi – 9.8 * t
37.6 -34.72 = 9.8 * t
t=0.293 or 0.3 s