/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 A cement block accidentally fall... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 61

A cement block accidentally falls from rest from the ledge of a \(53.0-\mathrm{m}\) -high building. When the block is 14.0 \(\mathrm{m}\) above the ground, a man, 2.00 \(\mathrm{m}\) 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?

Short Answer

Expert verified
The man has about 1.57 seconds to move out of the way.

Step by step solution

01

Understanding the Problem

A cement block falls from a height of 53.0 m and is noticed 14.0 m above the ground by a man who is 2.00 m tall. We need to find out the time the man has before the block reaches his height.
02

Determine Falling Distance

The block is noticed at 14.0 m above the ground. The man's eye level is at a height equivalent to the block's highest point (2.00 m above the ground). Therefore, the block has 12.0 m (14.0 m - 2.0 m) to fall before reaching the man's height.
03

Using Kinematics Equation

We use the kinematic equation: \[ d = \frac{1}{2}gt^2 \]where \(d\) is the distance fallen (12.0 m), and \(g\) is the acceleration due to gravity (9.8 m/s虏). We'll solve for \(t\), the time it takes to fall this distance.
04

Solving for Time

Substituting the known values into the kinematics equation gives:\[ 12.0 = \frac{1}{2} \cdot 9.8 \cdot t^2 \]This simplifies to:\[ 12.0 = 4.9t^2 \]Next, divide both sides by 4.9:\[ t^2 = \frac{12.0}{4.9} \approx 2.45 \]Then find \(t\) by taking the square root:\[ t = \sqrt{2.45} \approx 1.57 \text{ seconds} \]
05

Conclusion

The man has approximately 1.57 seconds to move out of the way before the falling block reaches him.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free Fall
When an object is in free fall, it means it is falling under the influence of gravity alone, without any additional forces like air resistance acting on it. In this scenario, the cement block falls freely from a building ledge. Let's break down the scene:

1. **Starting conditions**: The block begins its descent from a height of 53 meters above the ground, and it starts from rest, meaning its initial velocity is 0.
2. **Observation point**: The block is noticed when it's 14 meters above the ground.

Free falling objects accelerate as they descend, picking up speed due to gravity, which moves us to our next concept, acceleration due to gravity.
Acceleration Due to Gravity
Gravity is a force that pulls objects towards each other, and on Earth, it gives objects a constant downward acceleration. The standard value for this acceleration, often symbolized as \( g \), is 9.8 m/s虏. This means that for every second of fall, an object's downward velocity increases by 9.8 meters per second.

This constant acceleration plays a vital role in determining how fast an object falls, and therefore, how quickly it covers distance. Using this knowledge, we can formulate and solve equations to predict the motion of the falling block. In our problem, \( g \) helps us calculate how long the block takes to fall the remaining 12 meters to the man's height.
Distance Calculation
Distance calculation in free fall scenarios is often handled using a kinematic equation that directly relates an object's fall distance to the time elapsed, given acceleration. For a freely falling object, this equation is:
\[ d = \frac{1}{2}gt^2 \]
where:
  • \( d \) is the distance fallen (12 meters from the observation point to the man's height).
  • \( g \) is the acceleration due to gravity (9.8 m/s虏).
  • \( t \) is the time of fall.

This equation allows us to determine how far an object has fallen over a specific time span under the sole influence of gravity. By knowing the values of \( g \) and \( d \), we plug them into the formula to find \( t \), the focus of our next section.
Time Calculation
Solving for time begins with rearranging our kinematic equation to solve for \( t \). We had:
\[ 12.0 = \frac{1}{2} imes 9.8 imes t^2 \]
First, simplify by calculating \( \frac{1}{2} \times 9.8 \), which gives 4.9, then divide both sides of the equation by this value:
\[ t^2 = \frac{12.0}{4.9} \approx 2.45 \]
To extract \( t \), take the square root of 2.45, yielding approximately 1.57 seconds. This calculated time represents the available duration for the man to move aside safely. Therefore, he has about 1.57 seconds to react before the cement block reaches him. This prompt calculation highlights how vital the understanding of kinematics is in real-world scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An astronaut on a distant planet wants to determine its acceleration due to gravity. The astronaut throws a rock straight up with a velocity of \(+15 \mathrm{m} / \mathrm{s}\) and measures a time of 20.0 \(\mathrm{s}\) before the rock returns to his hand. What is the acceleration (magnitude and direction) due to gravity on this planet?

A Boeing 747 Jumbo Jet has a length of 59.7 \(\mathrm{m}\) . The runway on which the plane lands intersects another runway. The width of the inter- section is 25.0 \(\mathrm{m}\) . The plane decelerates through the intersection at a rate of 5.70 \(\mathrm{m} / \mathrm{s}^{2}\) and clears it with a final speed of 45.0 \(\mathrm{m} / \mathrm{s}\) . How much time is needed for the plane to clear the intersection?

In a historical movie, two knights on horseback start from rest 88.0 \(\mathrm{m}\) apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.300 \(\mathrm{m} / \mathrm{s}^{2}\) , while Sir Alfred's has a magnitude of 0.200 \(\mathrm{m} / \mathrm{s}^{2}\) . Relative to Sir George's starting point, where do the knights collide?

Review Conceptual Example 7 as background for this problem. A car is traveling to the left, which is the negative direction. The direction of travel remains the same throughout this problem. The car's initial speed is \(27.0 \mathrm{m} / \mathrm{s},\) and during a 5.0 \(\mathrm{s}\) - interval, it changes to a final speed of (a) 29.0 \(\mathrm{m} / \mathrm{s}\) and \((\mathrm{b}) 23.0 \mathrm{m} / \mathrm{s}\) . In each case, find the acceleration ( magnitude and algebraic sign) and state whether or not the car is decelerating.

A golfer rides in a golf cart an average speed of 3.10 \(\mathrm{m} / \mathrm{s}\) for 28.0 s. She then gets out of the cart and starts walking at an average speed of 1.30 \(\mathrm{m} / \mathrm{s}\) . For how long (in seconds) must she walk if her average speed for the entire trip, riding and walking, is 1.80 \(\mathrm{m} / \mathrm{s} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Dynamics

Read Explanation

Famous Physicists

Read Explanation

Solid State Physics

Read Explanation

Engineering Physics

Read Explanation

Energy Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.