/*! 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 56 Shows the speed \(v\) versus hei... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Shows the speed \(v\) versus height \(y\) of a ball tossed directly upward, along a \(y\) axis. Distance \(d\) is \(0.40 \mathrm{~m}\).The speed at height \(y_{A}\) is \(v_{A} .\) The speed at height \(y_{B}\) is \(\frac{1}{3} v_{A} .\) What is speed \(v_{A}\) ?

Short Answer

Expert verified
The speed \(v_A\) is approximately 1.26 m/s.

Step by step solution

01

Understand the Problem

We have a ball thrown upward along the y-axis, and we are given speeds at two heights: one at height \(y_A\) and another at height \(y_B\). We need to find the initial speed \(v_A\) at height \(y_A\), knowing that at height \(y_B\), the speed is \(\frac{1}{3} v_A\) and the height difference \(d = y_B - y_A = 0.40\; \text{m}\).
02

Apply Energy Conservation

Since gravitational force is acting on the ball, energy conservation principles apply. The total mechanical energy (kinetic + potential) at any two points is the same: \(\frac{1}{2} m v_A^2 + m g y_A = \frac{1}{2} m \left( \frac{1}{3} v_A \right)^2 + m g y_B\) where \(g\) is the acceleration due to gravity.
03

Simplify the Energy Equation

By simplifying, we cancel the mass \(m\) as it is present in every term: \[\frac{1}{2} v_A^2 + g y_A = \frac{1}{2} \left( \frac{1}{3} v_A \right)^2 + g (y_A + 0.40)\]. This general equation allows us to substitute numbers to find \(v_A\).
04

Substitute and Solve for \(v_A\)

Substituting \(g = 9.8 \; \text{m/s}^2\) and \(d = 0.40\; m\), the equation becomes: \[\frac{1}{2} v_A^2 = \frac{1}{2} \left( \frac{1}{9} v_A^2 \right) + 0.40 \times 9.8\] which simplifies to \[\frac{44}{18} v_A^2 = 3.92\]. Solving for \(v_A\) gives us \(v_A^2 = \frac{3.92 \times 18}{44}\) and \(v_A = \sqrt{\frac{70.56}{44}}\).
05

Solve for the Exact Value

Evaluating the expression above: \(v_A = \sqrt{1.60} \approx 1.26 \; \text{m/s}.\)

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.

Kinetic Energy
Kinetic energy is the energy associated with an object's motion. It depends on both the mass of the object and its velocity. The faster an object moves, the greater its kinetic energy. This energy can be calculated using the formula:
  • \[ KE = \frac{1}{2} m v^2 \]
where \( m \) stands for the mass of the object and \( v \) represents its velocity. When a ball is thrown upwards, like in the original exercise, its kinetic energy is highest at the start when the velocity is maximum. As the ball rises, kinetic energy decreases because speed decreases until it reaches zero at the highest point of its trajectory. Understanding kinetic energy helps us see how energy is conserved and transformed from kinetic to potential energy in the motion of a ball.
Potential Energy
Potential energy refers to the energy stored in an object due to its position in a force field, commonly a gravitational field when dealing with height. In the context of a ball tossed upward, potential energy increases as the ball rises in height. This is because the ball is moving against gravitational pull.
  • The formula for gravitational potential energy is \( PE = mgh \),where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.
As the ball gains height, it stores more potential energy at the expense of kinetic energy, because of the energy conservation principle. At the ball's highest point, its potential energy is maximum and kinetic energy is minimum. Understanding this concept is crucial for solving problems related to energy transformation in vertical motion.
Gravitational Force
Gravitational force is an attractive force that acts between any two masses. It is particularly significant in cases involving objects on or near Earth, like the tossed ball. This force pulls the ball downward, opposing its upward motion. The acceleration caused by gravitational force near the Earth’s surface is denoted as \( g \) and approximately equals \( 9.8 \, \text{m/s}^2 \).
  • This constant acceleration is integral to calculations involving energy conservation and kinetics.
  • It is what converts kinetic energy to potential energy as the ball moves upward, and vice versa when it falls back.
Gravitational force is essential in understanding how potential and kinetic energy interchange, providing a basis for calculating changes in speed and energy during the ball's flight.
Initial Speed Calculation
Initial speed is crucial to determine the motion path and speed at various points of the projectile's journey. To find the initial speed in energy conservation scenarios, like the exercise given, we use the principle that the total mechanical energy remains constant — the sum of kinetic and potential energy is constant at any point during the motion.
  • The equation: \[ \frac{1}{2} m v_A^2 + m g y_A = \frac{1}{2} m \left( \frac{1}{3} v_A \right)^2 + m g (y_A + 0.40) \]shows this condition, with initial speed \( v_A \) at height \( y_A \).
  • Rearranging and solving such equations lets us find unknown variables, like \( v_A \), by focusing on energy conservation and known values like gravitational acceleration and height difference.
For the example exercise, solving these steps and rearranging gives the initial speed, informing us of how variables like height and speed interdepend. It emphasizes the underlying law that energy cannot be created or destroyed, only changed from one form to another.

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

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 \(3.0\) and \(2.0\), respectively. (c) At what time does the particle reach its maximum positive \(x\) position? From \(t=0.0 \mathrm{~s}\) to \(t=4.0 \mathrm{~s},(\mathrm{~d})\) what distance does the particle move and (e) what is its displacement? Find its velocity at times (f) \(1.0 \mathrm{~s},(\mathrm{~g}) 2.0 \mathrm{~s},(\mathrm{~h}) 3.0 \mathrm{~s}\), and (i) \(4.0 \mathrm{~s}\). Find its acceleration at times (j) \(1.0 \mathrm{~s},(\mathrm{k}) 2.0 \mathrm{~s},(1) 3.0 \mathrm{~s}\), and \((\mathrm{m}) 4.0 \mathrm{~s}\).

A lead ball is dropped in a lake from a diving board \(5.20 \mathrm{~m}\) above the water. It hits the water with a certain velocity and then sinks to the bottom with this same constant velocity. It reaches the bottom \(4.80 \mathrm{~s}\) after it is dropped. (a) How deep is the lake? What are the (b) magnitude and (c) direction (up or down) of the average velocity of the ball for the entire fall? Suppose that all the water is drained from the lake. The ball is now thrown from the diving board so that it again reaches the bottom in \(4.80 \mathrm{~s}\). What are the (d) magnitude and (e) direction of the initial velocity of the ball?

A parachutist bails out and freely falls \(50 \mathrm{~m}\). Then the parachute opens, and thereafter she decelerates at \(2.0 \mathrm{~m} / \mathrm{s}^{2}\). She reaches the ground with a speed of \(3.0 \mathrm{~m} / \mathrm{s}\). (a) How long is the parachutist in the air? (b) At what height does the fall begin?

At the instant the traffic light turns green, an automobile starts with a constant acceleration \(a\) of \(2.2 \mathrm{~m} / \mathrm{s}^{2} .\) At the same instant a truck, traveling with a constant speed of \(9.5 \mathrm{~m} / \mathrm{s}\), overtakes and passes the automobile. (a) How far beyond the traffic signal will the automobile overtake the truck? (b) How fast will the automobile be traveling at that instant?

A hot rod can accelerate from 0 to \(60 \mathrm{~km} / \mathrm{h}\) in \(5.4 \mathrm{~s}\). (a) What is its average acceleration, in \(\mathrm{m} / \mathrm{s}^{2}\), during this time? (b) How far will it travel during the \(5.4 \mathrm{~s}\), assuming its acceleration is constant? (c) From rest, how much time would it require to go a distance of \(0.25 \mathrm{~km}\) if its acceleration could be maintained at the value in (a)?

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.