/*! 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 33 A stone is thrown from the top o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 33

A stone is thrown from the top of a building with an initial velocity of \(20 \mathrm{~m} / \mathrm{s}\) downward. The top of the building is \(60 \mathrm{~m}\) above the ground. How much time elapses between the instant of release and the instant of impact with the ground?

Short Answer

Expert verified
The stone hits the ground after approximately 2.01 seconds.

Step by step solution

01

Identify Given Variables

We have the initial velocity \( v_0 = 20 \, \text{m/s} \) (downward), the height of the building \( h = 60 \, \text{m} \), and the acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \). We need to calculate the time \( t \) it takes for the stone to hit the ground.
02

Apply the Kinematic Equation

Use the equation for motion with constant acceleration: \( h = v_0 t + \frac{1}{2} g t^2 \). Here, \( h = 60 \, \text{m} \), \( v_0 = 20 \, \text{m/s} \), and \( g = 9.8 \, \text{m/s}^2 \). Substitute these values into the equation.
03

Rearrange and Simplify the Equation

The equation becomes \( 60 = 20t + \frac{1}{2} \times 9.8 \times t^2 \). Simplify this to \( 60 = 20t + 4.9t^2 \).
04

Solve the Quadratic Equation

To find \( t \), rearrange \( 4.9t^2 + 20t - 60 = 0 \). Use the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4.9 \), \( b = 20 \), and \( c = -60 \).
05

Calculate the Discriminant

Calculate the discriminant: \( b^2 - 4ac = 20^2 - 4 \times 4.9 \times (-60) \). This equals \( 400 + 1176 = 1576 \).
06

Find the Roots of the Quadratic Equation

Substitute the discriminant into the quadratic formula to find \( t \): \( t = \frac{-20 \pm \sqrt{1576}}{2 \times 4.9} \). Calculate both possible values to find the positive root, as time cannot be negative.
07

Select the Positive Root

Evaluate the roots and select the positive value: \( t = \frac{-20 + 39.7}{9.8} \approx 2.01 \). The other root will be negative or physically irrelevant.

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.

Quadratic Equation
In kinematics, a quadratic equation often appears when solving for variables like time or displacement, particularly under constant acceleration. These equations take the form:\[ ax^2 + bx + c = 0 \]Here, \(a\), \(b\), and \(c\) are coefficients. When analyzing motion, these coefficients are related to factors like initial velocity and acceleration.
  • Quadratic equations can have two solutions, known as roots.
  • These are found using the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
  • The expression under the square root \( b^2 - 4ac \) is called the discriminant, which determines the number and type of solutions.
In the context of the stone being thrown, solving this quadratic equation helps us find the time at which it impacts the ground.
Initial Velocity
Initial velocity (\( v_0 \)) describes how fast something moves at the start of our observation. It sets a baseline for calculating future motion. In this scenario, the stone starts with an initial velocity of \(20 \,\mathrm{m/s}\). This velocity is directed downward. The initial velocity:
  • Directly influences the motion's future position and speed.
  • Is a key value in kinematic equations, affecting time and displacement calculations.
  • Can be positive or negative based on its direction of motion.
A positive initial velocity typically means the object starts moving in the reference direction, while negative signals the opposite.
Acceleration Due to Gravity
Gravity influences every object on Earth's surface by exerting a downward force. Acceleration due to gravity, denoted as \(g\), has a standard value of approximately \(9.8 \, \mathrm{m/s}^2\). This acceleration acts continuously on freely moving objects, increasing their velocity over time.In our stone-throwing problem, gravity's role is crucial:
  • It continuously accelerates the stone downward at \(9.8 \, \mathrm{m/s}^2\).
  • Over time, gravity increases the stone's speed, impacting calculations of displacement and time.
  • This uniform acceleration is a fundamental part of the kinematic equations used to calculate motion.
Understanding gravity's consistent pull is essential for accurately applying these equations.
Displacement
Displacement refers to an object's change in position. It is a vector quantity, meaning it has both magnitude and direction. In the context of our exercise, displacement is the change in position of the stone from the top of the building to the ground.
  • It is affected by both initial velocity and acceleration.
  • Displacement is part of the kinematic equation: \( h = v_0 t + \frac{1}{2} g t^2 \).
  • The equation sums initial motion and acceleration effects over time.
In practical terms, displacement helps us understand how far and in what direction the stone travels from its starting point.

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

In a particle accelerator, an electron enters a region in which it accelerates uniformly in a straight line from a speed of \(4.00 \times 10^{5}\) \(\mathrm{m} / \mathrm{s}\) to a speed of \(6.00 \times 10^{7} \mathrm{~m} / \mathrm{s}\) in a distance of \(3.00 \mathrm{~cm}\). For what time interval does the electron accelerate?

A hot-air balloon is ascending at the rate of \(12 \mathrm{~m} / \mathrm{s}\) and is \(80 \mathrm{~m}\) above the ground when a package is dropped over the side. (a) How long does the package take to reach the ground? (b) With what speed does it hit the ground?

A man releases a stone at the top edge of a tower. During the last second of its travel, the stone falls through a distance of \((9 / 25) H\), where \(H\) is the tower's height. Find \(H\).

The position of a particle moving along the \(x\) axis depends on the time according to the equation \(x=c t^{2}-b t^{3}\), where \(x\) is in meters and \(t\) in seconds. What are the units of (a) constant \(c\) and (b) constant \(b\) ? Let their numerical values be \(4.0\) and \(2.0\), respectively. (c) At what time does the particle reach its maximum positive \(x\) position? From \(t=0.0 \mathrm{~s}\) to \(t=4.0 \mathrm{~s}\), (d) what distance does the particle move and (e) what is its displacement? Find its velocity at times (f) \(1.0 \mathrm{~s}\), (g) \(2.0 \mathrm{~s}\), (h) \(3.0 \mathrm{~s}\), and (i) \(4.0 \mathrm{~s}\). Find its acceleration at times (j) \(1.0 \mathrm{~s}\), (k) \(2.0 \mathrm{~s}\), (l) \(3.0 \mathrm{~s}\), and \((\mathrm{m}) 4.0 \mathrm{~s}\).

In \(1 \mathrm{~km}\) races, runner 1 on track 1 (with time \(2 \mathrm{~min}, 27.95 \mathrm{~s}\) ) appears to be faster than runner 2 on track 2 ( \(2 \mathrm{~min}, 28.15 \mathrm{~s}\) ). However, length \(L_{2}\) of track 2 might be slightly greater than length \(L_{1}\) of track 1. How large can \(L_{2}-L_{1}\) be for us still to conclude that runner 1 is faster?
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Fluids

Read Explanation

Astrophysics

Read Explanation

Electricity

Read Explanation

Physics of Motion

Read Explanation

Radiation

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.