College

Thank you for visiting A cement block accidentally falls from rest from the ledge of a 51 1 meter high building When the block is 13 4 meters above. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!

A cement block accidentally falls from rest from the ledge of a 51.1-meter-high building. When the block is 13.4 meters above the ground, a man, 1.70 meters 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 :

Final answer:

The man has approximately 1.46 seconds to get out of the way.

Explanation:

To find the maximum time the man has to get out of the way, we can first calculate the time it takes for the block to fall from 51.1 m to 13.4 m using the equation: h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. Rearranging the equation gives us: t = sqrt(2h/g). Plugging in the values, we find that it takes approximately 1.46 seconds for the block to fall from 51.1 m to 13.4 m.

Next, we can calculate the time it takes for the block to fall from 13.4 m to the ground using the same equation. Plugging in the new height of 13.4 m, we find that it takes approximately 0.53 seconds to fall this distance.

Therefore, the maximum time the man has to get out of the way is the time it takes for the block to fall from 51.1 m to 13.4 m, which is approximately 1.46 seconds.

Learn more about Falling block here:

https://brainly.com/question/32272111

#SPJ3

Thank you for reading the article A cement block accidentally falls from rest from the ledge of a 51 1 meter high building When the block is 13 4 meters above. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!

Rewritten by : Jeany

Answer:

The man has, at most, 0.4 secs to get out of the way

Explanation:

First, we will determine, the time the block takes to reach height 13.4 m above the ground.

Also, we will determine the time it will take to reach the man, if he does not notice it.

Then, the difference in the two times will be the time the man has to get out of the way.

From the formula

[tex]h =ut - \frac{1}{2}gt^{2}[/tex]

Where [tex]h[/tex] is height

[tex]u[/tex] is the initial velocity

[tex]t[/tex] is time

and [tex]g[/tex] is acceleration due to gravity ( Take [tex]g =[/tex] 9.8 m/s²)

Let the time the block takes to reach height 13.4 m be [tex]t_{1}[/tex]

The cement block falls from a 51.1 m high building, then by the time it reaches a height 13.4 m above the ground, the displacement is

13.4 m - 51.1 m = - 37.7 m

The is the height the block has fallen from by the time it gets to a height 13.4 m above the ground

Then,

[tex]h =ut - \frac{1}{2}gt^{2}[/tex]

[tex]-37.7 = (0)(t_{1}) - \frac{1}{2}(9.8)t_{1}^{2}[/tex]

(NOTE: [tex]u[/tex] = 0 m/s because the cement block falls from rest)

[tex]37.7 = 4.9t_{1}^{2}[/tex]

[tex]t_{1}^{2}= \frac{37.7}{4.9}[/tex]

[tex]t_{1}= \sqrt{7.69} \\[/tex]

[tex]t_{1}= 2.77 secs[/tex]

Hence, It takes the block 2.77 secs to reach height 13.4 m

Also,

Let the time it will take to reach the man be [tex]t_{2}[/tex]

The man is 1.70 m tall, hence, to reach the man, the cement block would have reached a height ( 1.70 m - 51.1 m) = - 49.4 m

Then,

[tex]h =ut - \frac{1}{2}gt^{2}[/tex]

[tex]-49.4 = (0)(t_{2}) - \frac{1}{2}(9.8)t_{2}^{2}[/tex]

[tex]49.4 = 4.9t_{2}^{2}[/tex]

[tex]t_{2}^{2} = \frac{49.4}{4.9}[/tex]

[tex]t_{2}= \sqrt{10.08}[/tex]

[tex]t_{2}= 3.17 secs[/tex]

Hence, it will take 3.17 secs to reach the man

Now, for the time the man has to get out of the way, that is

[tex]t_{2} - t_{1}[/tex] = 3.17 secs - 2.77 secs

= 0.4 secs

Hence, the man has, at most, 0.4 secs to get out of the way

Other Questions