/*! 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] | 91Ó°ÊÓ

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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 approximately 1.69 seconds to move.

Step by step solution

01

Understanding the Problem

We are given that a cement block falls from a 53.0 m height and we need to find the time it takes for the block to go from 14.0 m above the ground to the ground, so the man has at most this time to get out of the way.
02

Identify Known Values

Initial velocity (\(v_i\)) is 0 m/s since it falls from rest. The distance from 14.0 m to 0 m (ground level) is 14.0 m. Gravitational acceleration (\(g\)) is 9.8 m/s².
03

Applying the Kinematic Equation

The kinematic equation \(d = v_i t + \frac{1}{2}at^2\) can be used. Since the block falls from rest, \(d = \frac{1}{2}gt^2\). Here, \(d = 14.0 \) m and \(a = g = 9.8 \) m/s².
04

Solve for Time

Rearrange the equation \(14.0 = \frac{1}{2} \cdot 9.8 \cdot t^2\). This simplifies to \(14.0 = 4.9t^2\). Divide both sides by 4.9 to get \(t^2 = \frac{14.0}{4.9}\), which gives \(t^2 \approx 2.857\). Taking the square root, \(t \approx \sqrt{2.857} \approx 1.69\) seconds.

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Key Concepts

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

Gravitational Acceleration
Gravitational acceleration is a key concept in kinematics and plays a significant role in various motions we observe daily. On Earth, gravitational acceleration, denoted by \( g \), is approximately \( 9.8 \text{ m/s}^2 \). This value indicates how fast an object accelerates when it is falling freely under the influence of gravity alone. In other words, any object dropped from rest will increase its speed by \( 9.8 \text{ m/s} \) for every second it is falling.
- This acceleration is constant and acts downward towards the center of the Earth.
- It's essential to remember gravitational acceleration when solving problems involving free-falling objects.
In kinematic equations, \( g \) is often used as the acceleration component, particularly in situations where air resistance is negligible. In our scenario with the cement block, \( g \) helps us determine the time it takes for the block to travel from 14.0 meters above the ground to hitting the ground. Understanding gravitational acceleration is foundational for further calculations in kinematics.
Initial Velocity
Initial velocity is the speed of an object before any forces are applied to change that speed. In the problem, initial velocity is crucial as it determines how the motion of the cement block begins. Since the block is initially at rest, this sets the initial velocity \( v_i \) to \( 0 \text{ m/s} \).
  • "At rest" means there is no initial motion at the time the block starts falling.
  • Knowing the initial velocity allows us to simplify calculations in kinematic equations.
For cases like free-fall, where objects begin from a stationary position, the initial velocity is zero. This simplification is vital because it allows us to use a particular form of the kinematic equation, highlighting the relationship between distance and time without the need for initial velocity considerations. In this context, recognizing that the initial velocity is zero helps streamline solving the problem, focusing on the straightforward impact of gravitational forces alone.
Kinematic Equation
Kinematic equations are mathematical formulas used to describe the motion of objects. They relate the five key components of motion: displacement, initial velocity, final velocity, acceleration, and time. One commonly used kinematic equation in free-fall problems is: \[d = v_i t + \frac{1}{2} a t^2\]In the exercise, since the block falls from rest, the initial velocity \( v_i = 0 \), simplifying our equation to: \[d = \frac{1}{2} g t^2\]Here's how we use this equation to find the time it takes for the cement block to fall 14.0 meters:- Substitute \( d = 14.0 \text{ m} \) and \( a = g = 9.8 \text{ m/s}^2 \) into the equation.- Solve for time \( t \) by first isolating \( t^2 \): \[ 14.0 = \frac{1}{2} \cdot 9.8 \cdot t^2 \] \[ 14.0 = 4.9 t^2 \] \[ t^2 = \frac{14.0}{4.9} \]- Taking the square root results in \( t \approx 1.69 \) seconds.This example demonstrates how, by knowing just a few values, we can find unknowns related to an object's motion using the kinematic equation. These equations are crucial tools in physics to predict and understand various types of motion in a structured way.

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Most popular questions from this chapter

A hot-air balloon is rising upward with a constant speed of\(2.50 \mathrm{m} / \mathrm{s} .\) When the balloon is \(3.00 \mathrm{m}\) above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?

An Australian emu is running due north in a straight line at a speed of \(13.0 \mathrm{m} / \mathrm{s}\) and slows down to a speed of \(10.6 \mathrm{m} / \mathrm{s}\) in \(4.0 \mathrm{s}\). (a) What is the direction of the bird's acceleration? (b) Assuming that the acceleration remains the same, what is the bird's velocity after an additional \(2.0 \mathrm{s}\) has elapsed?

A locomotive is accelerating at \(1.6 \mathrm{m} / \mathrm{s}^{2} .\) It passes through a \(20.0-\mathrm{m}-\) wide crossing in a time of \(2.4 \mathrm{s} .\) After the locomotive leaves the crossing, how much time is required until its speed reaches \(32 \mathrm{m} / \mathrm{s} ?\)

Electrons move through a certain electric circuit at an average speed of \(1.1 \times 10^{-2} \mathrm{m} / \mathrm{s} .\) How long (in minutes) does it take an electron to traverse \(1.5 \mathrm{m}\) of wire in the filament of a light bulb?

A tourist being chased by an angry bear is running in a straight line toward his car at a speed of \(4.0 \mathrm{m} / \mathrm{s}\). The car is a distance \(d\) away. The bear is \(26 \mathrm{m}\) behind the tourist and running at \(6.0 \mathrm{m} / \mathrm{s}\). The tourist reaches the car safely. What is the maximum possible value for \(d\) ?

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