Question

a cement block accidentally falls from rest from the ledge of a 80.6-m-high building. When the block is 10.8 m above the ground, a man, 2.10 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

Answer:

0.229 seconds

Explanation:

Given:

y₀ = 80.6 m

v₀ = 0 m/s

a = -9.8 m/s²

We need to find the difference in times when y = 10.8 m and y = 2.10 m.

When y = 10.8 m:

y = y₀ + v₀ t + ½ at²

10.8 = 80.6 + (0) t + ½ (-9.8) t²

10.8 = 80.6 − 4.9 t²

4.9 t² = 69.8

t = 3.774

When y = 2.10 m:

y = y₀ + v₀ t + ½ at²

2.10 = 80.6 + (0) t + ½ (-9.8) t²

2.10 = 80.6 − 4.9 t²

4.9 t² = 78.5

t = 4.003

The difference is:

4.003 − 3.774 = 0.229

The man has 0.229 seconds to get out of the way.