/*! 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 42 A brick is dropped (zero initial... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 2.50 s. You may ignore air resistance, so the brick is in free fall. (a) How tall, in meters, is the building? (b) What is the magnitude of the brick's velocity just before it reaches the ground? (c) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion of the brick.

Short Answer

Expert verified
(a) 30.66 m, (b) 24.53 m/s, (c) Sketch the graphs as described.

Step by step solution

01

Determine the acceleration

Since the brick is in free fall, the only acceleration it experiences is due to gravity. This acceleration is constant and denoted by \( g \), with a value of approximately \( 9.81 \text{ m/s}^2 \) downward.
02

Use the kinematic equation to find the height

To determine the height of the building, use the kinematic equation: \[ y = v_i t + \frac{1}{2} a t^2 \]Since the initial velocity \( v_i \) is 0 (dropped from rest), the equation simplifies to:\[ y = \frac{1}{2} g t^2 \]Substitute \( g = 9.81 \text{ m/s}^2 \) and \( t = 2.5 \text{ s} \):\[ y = \frac{1}{2} \times 9.81 \times (2.5)^2 = 30.66 \text{ m} \]Thus, the height of the building is approximately 30.66 meters.
03

Find the final velocity

Use the kinematic equation: \[ v = v_i + a t \]Again, since \( v_i = 0 \):\[ v = g t \]Substituting \( g = 9.81 \text{ m/s}^2 \) and \( t = 2.5 \text{ s} \):\[ v = 9.81 \times 2.5 = 24.525 \text{ m/s} \]The magnitude of the velocity just before it reaches the ground is approximately 24.53 m/s.
04

Sketch the graphs

1. **Acceleration vs. Time \((a_y-t)\)**: The graph is a horizontal line at \( a = 9.81 \text{ m/s}^2 \), indicating constant acceleration throughout the fall.2. **Velocity vs. Time \((v_y-t)\)**: This graph is a straight line starting from the origin (0,0) and ending at \( (2.5, 24.53) \). The slope of the line is equal to the acceleration due to gravity.3. **Height vs. Time \((y-t)\)**: This graph is a curve starting at the maximum height (\( y = 30.66 \)) and decreasing to 0, showing a parabolic descent as it approaches the ground.

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 91Ó°ÊÓ!

Key Concepts

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

Free Fall
In the context of physics, free fall describes the motion of an object moving solely under the influence of gravity. When objects are in free fall, they experience an acceleration toward the center of the Earth due to gravity, which is approximately 9.81 m/s². This means that for every second an object is in free fall, its velocity increases by 9.81 m/s. It's important to note that free fall assumes no air resistance.

When a brick is dropped from a building, it starts from rest, meaning its initial velocity is zero. As the brick falls under the influence of gravity, it speeds up, reaching a higher velocity just before it hits the ground. Since we're dealing with a downward force, the acceleration is constant and directs towards the earth.
Kinematic Equations
Kinematic equations are fundamental in understanding motion, especially in problems involving constant acceleration like free fall. These equations relate the variables of motion: displacement, velocity, acceleration, and time. They help us find unknowns when certain variables are given.

In the exercise, the equation \( y = v_i t + \frac{1}{2} a t^2 \) was used to find the height from which the brick was dropped. Because the brick is initially at rest, the equation simplifies to \( y = \frac{1}{2} g t^2 \). This relationship shows how displacement changes over time under constant acceleration. Similarly, to find velocity just before impact, the equation \( v = v_i + a t \) comes into play. With an initial velocity of zero, this simplifies to \( v = g t \). These equations become powerful tools for solving real-life motion problems.
Velocity-Time Graph
A velocity-time graph provides insights into how an object's velocity changes over time. For an object in free fall, the velocity-time graph will show a straight line with a positive slope. This straight line reflects the constant acceleration due to gravity.

In our problem, the graph starts from the origin because the initial velocity is zero. As time progresses, the brick accelerates, and its velocity increases at a constant rate. By the time the brick hits the ground at 2.5 seconds, the graph reaches a velocity of approximately 24.53 m/s. The slope of this line is equal to the acceleration due to gravity (9.81 m/s²), visually demonstrating how gravity affects the brick's velocity.
Acceleration-Time Graph
An acceleration-time graph illustrates how acceleration changes with respect to time. In the scenario of free fall, the acceleration remains constant. Thus, the graph is depicted as a horizontal line. For the brick in the exercise, the acceleration-time graph shows a constant acceleration of 9.81 m/s², horizontally stretching across the timeline for the whole fall duration. This constancy indicates that gravity's influence doesn’t vary while the object is in motion, provided there's no air resistance. Such a graph is pivotal in understanding the uniform nature of gravitational acceleration during free fall.
Displacement-Time Graph
A displacement-time graph portrays how the position or height of an object changes over time. In free fall, this graph typically takes the shape of a parabola.

At the start, the brick is at a height of approximately 30.66 meters. Over time, as the brick falls, its displacement from the starting point increases rapidly. This increasing rate of change creates a parabolic curve on the graph. It begins at the highest point and drops to zero at 2.5 seconds when the brick strikes the ground. This curve visually demonstrates the increasing speed of the object as it descends, highlighting the acceleration's effect on motion.

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

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 \(\mathrm{m} / \mathrm{s}\) . Air resistance may be ignored. (a) At what time after being ejected is the boulder moving at 20.0 \(\mathrm{m} / \mathrm{s}\) upward? (b) At what time is it moving at 20.0 \(\mathrm{m} / \mathrm{s}\) downward? (c) When is the displacement of the boulder from its initial position zero? (d) When is the velocity of the boulder zero? (e) What are the magnitude and direction of the acceleration while the boulder is (i) moving upward? (ii) Moving downward? (iii) At the highest point? (f) Sketch \(a_{y}-t, v_{y}-t,\) and \(y-t\) graphs for the motion.

Sam heaves a \(16-\) - Ib shot straight upward, giving it a constant upward accleration from rest of 45.0 \(\mathrm{m} / \mathrm{s}^{2}\) for 64.0 \(\mathrm{cm} .\) He releases it 2.20 \(\mathrm{m}\) above the ground. You may ignore air resistance. \((\mathrm{a})\) What is the speed of the shot when Sam releases it? (b) How high above the ground does it go? (c) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 \(\mathrm{m}\) above the ground?

A turtle crawls along a straight line, which we will call the \(x\) -axis with the positive dircction to the right. The equation for the turtle's position as a function of time is \(x(t)=50.0 \mathrm{cm}+\) \((2.00 \mathrm{cm} / \mathrm{s}) t-\left(0.0625 \mathrm{cm} / \mathrm{s}^{2}\right) t^{2},(\mathrm{a})\) Find the turte's initial velocity, initial position, and initial acceleration. (b) At what time \(t\) is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times \(t\) is the turtle a distance of 10.0 \(\mathrm{cm}\) from its starting point? What is the velocity (magnitude and direction) of the turtle at each of these times? (e) Sketch graphs of \(x\) versus \(t, v_{x}\) versus \(t,\) and \(a_{x}\) versus \(t\) for the time interval \(t=0\) to \(t=40 \mathrm{s}\)

A ball starts from rest and rolls down a hill with uniform acceleration, traveling 150 \(\mathrm{m}\) during the second 5.0 \(\mathrm{s}\) of its motion. How far did it roll during the first 5.0 \(\mathrm{s}\) of motion?

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof 1.00 s later. You may ignore air resistance. (a) If the height of the building is \(20.0 \mathrm{m},\) what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the position of each ball as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_{0}\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time (i) if \(v_{0}\) is 6.0 \(\mathrm{m} / \mathrm{s}\) and (ii) if \(v_{0}\) is 9.5 \(\mathrm{m} / \mathrm{s} ?(\mathrm{c})\) If \(v_{0}\) is greater than some value \(v_{\max }\) a value of \(h\) does not exist that allows both balls to hit the ground at the same time. Solve for \(v_{\text { max. The value }} v_{\text { max }}\) has a simple physical interpretation. What is it? (d) If \(v_{0}\) is less than some value \(v_{\text { min }}\) a value of \(h\) does not exist that allows both balls to hit the ground at the same time. Solve for \(v_{\text { main }}\) The value \(v_{\text { min }}\) also has a simple physical interpretation. What is it?

See all solutions

Recommended explanations on Physics Textbooks

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.