A cement block accidentally falls from rest from the ledge of a 53.0-m-high building. When the block is 14.0 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?

Question

Alison Lyons

Physics

23 October, 18:57

A cement block accidentally falls from rest from the ledge of a 53.0-m-high building. When the block is 14.0 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?

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Answers (1)

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Bryanna · Answer 1

0.405 seconds

Explanation:

Consider the amount of time it takes the block to fall from 53 m up to 14 m above the ground; then consider the amount of time it takes the block to fall from 53 m up to 2 m above the ground.

First, d = (1/2) gt^2 or t = (2 d / g) ^1/2

= (2 * 39 / 9.8) ^1/2 = 2.8212 seconds

Then, to fall from 53 down to 2 meters ...

d = (1/2) gt^2 or t = (2 d / g) ^1/2

= (2 * 51 / 9.8) ^1/2 = 3.2262 seconds

So the amount of time it takes for the block to fall from 14 m upto 2 m above the ground

3.2262 - 2.8212 = 0.405 seconds

this is how much time there is from when the man sees the block until it hits him. Not much time ...