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

Physics

BatMan

6 September, 18:14

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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Jaylin Trujillo · 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 ...