/*! 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 10 A stone of mass \(m\) is thrown ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A stone of mass \(m\) is thrown upward at an angle \(\theta\) above the horizontal and feels no appreciable air resistance. Use conservation of energy to show that at its highest point, it is a distance \(v_{0}^{2}\left(\sin ^{2} \theta\right) / 2 g\) above the point where it was launched. (Hint: \(v_{0}^{2}=v_{\mathrm{ax}}^{2}+v_{\mathrm{oy}}^{2} . )\)

Short Answer

Expert verified
The stone reaches a height of \( \frac{v_0^2 \sin^2 \theta}{2g} \) at its highest point.

Step by step solution

01

Identify Initial Energies

To use the conservation of energy, identify the initial kinetic energy (KE) of the stone. The initial speed is given by \( v_0 \), and the kinetic energy is calculated as \( KE_i = \frac{1}{2}mv_0^2 \).The initial potential energy (PE) is zero because the stone starts from the launch point, which we assume to be the reference level where potential energy is zero. Thus, \( PE_i = 0 \).
02

Break Down Initial Velocity into Components

The initial velocity \( v_0 \) can be broken into horizontal and vertical components using the angle \( \theta \).- Horizontal component: \( v_{0x} = v_0 \cos \theta \)- Vertical component: \( v_{0y} = v_0 \sin \theta \)
03

Determine Mechanical Energy at the Highest Point

At the highest point, the vertical component of the velocity is zero \( (v_{y} = 0) \), but the horizontal component remains, so:- Horizontal velocity at the top: \( v_{x} = v_0 \cos \theta \)Using conservation of mechanical energy, the mechanical energy at the top consists of the horizontal kinetic energy alone, since the vertical kinetic energy is zero.Thus, \( KE_t = \frac{1}{2}mv_{x}^2 = \frac{1}{2}m(v_0 \cos \theta)^2. \)The potential energy at the top is what we need to determine.
04

Apply Conservation of Energy

According to conservation of energy:Initial total energy = Total energy at the highest point.\[ KE_i + PE_i = KE_t + PE_t. \]Substituting the known values,\[ \frac{1}{2}mv_0^2 = \frac{1}{2}m(v_0 \cos \theta)^2 + mgh. \]Solve for \( h \), the height:\[ mv_0^2 - m(v_0 \cos \theta)^2 = 2mgh. \]
05

Solve for Height

Simplify the equation from Step 4:\[ mv_0^2 - mv_0^2 \cos^2 \theta = 2mgh \]\[ mv_0^2 (1 - \cos^2 \theta) = 2mgh \]\(1 - \cos^2 \theta = \sin^2 \theta \), so:\[ mv_0^2 \sin^2 \theta = 2mgh \].Solve for \( h \):\[ h = \frac{v_0^2 \sin^2 \theta}{2g}. \]
06

Conclusion

We have shown that at its highest point, the stone reaches a height of \( \frac{v_0^2 \sin^2 \theta}{2g} \) above the launch point, as required by the problem statement.

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.

Projectile Motion
Projectile motion is a fascinating concept in physics. It describes the motion of an object that is thrown or projected into the air, subject to only the acceleration of gravity. Such motion occurs in a curved path, and this path is known as a trajectory. For a projectile launched at an angle \( \theta \) to the horizontal, its initial velocity \( v_0 \) can be split into two components:
  • The horizontal component \( v_{0x} = v_0 \cos \theta\) remains constant throughout the motion, as there is no horizontal acceleration.
  • The vertical component \(v_{0y} = v_0 \sin \theta\) decreases upward until the highest point and then increases in the downward direction due to gravity.
As the stone reaches its highest point in this problem, the vertical velocity is zero, which allows us to find the maximum height using energy principles.
Kinetic Energy
Kinetic energy (KE) is the energy possessed by an object due to its motion. The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the object.
In the scenario of projectile motion, the stone initially has kinetic energy due to its initial speed \( v_0 \). As it moves upwards, its kinetic energy rearranges between its horizontal and vertical motion.
At the top of its path, its vertical kinetic energy is reduced to zero, while the horizontal component \( v_{x} \) maintains kinetic energy. The conserved nature of kinetic energy in the absence of non-conservative forces, like friction or air resistance, is key in analyzing projectile motion.
Potential Energy
Potential energy (PE) depends on the position of an object within a gravitational field. For an object of height \( h \) above a reference point, it is given by the equation \( PE = mgh \).
At the launch, the stone's potential energy is zero at the reference level. However, when the stone reaches its maximum height, the potential energy is at its peak, as all the initial energy has transformed.
The gain in height equates to the conversion of some initial kinetic energy to potential energy. Using energy conservation, we find the total height \( h \) reached by solving \( h = \frac{v_0^2 \sin^2 \theta}{2g} \), illustrating energy transformation from kinetic to potential energy.
Mechanics
Mechanics, a core branch of physics, deals with the motion of objects and the forces acting upon them. It is divided into classical mechanics, which includes the study of projectile motion.
This problem of projecting a stone explores the principles of classical mechanics by analyzing forces, velocities, and energy types. It highlights the conservation laws, which state that energy cannot be created or destroyed but only transformed from one form to another.
The principles of mechanics enable us to use formulas related to kinetic and potential energy effectively to predict heights, velocities, and trajectories of projectiles.
Physics Education
Physics education aims to introduce learners to concepts that explain the natural world. Using real-world problems, such as projectile motion, makes complex theories more engaging and understandable.
When students solve problems involving conservation of energy, they learn to apply theoretical knowledge in practical scenarios, enhancing their problem-solving skills. Understanding these foundational topics prepares learners for more advanced studies in physics and other scientific fields.
Educational platforms often offer step-by-step solutions to exercises, helping to bridge the gap between theory and practical application. Thorough explanations of each step can help clarify the interconnectedness of concepts like kinetic and potential energy within the broader context of physics.

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 truck loading station at a post office, a small 0.200 \(\mathrm{kg}\) package is released from rest at point \(A\) on a track that is one-quarter of a circle with radius 1.60 \(\mathrm{m}(\mathrm{Fig} .7 .39) .\) The size of the package is much less than \(1.60 \mathrm{m},\) so the package can be treated as a particle. It slides down the track and reaches point \(B\) with a speed of 4.80 \(\mathrm{m} / \mathrm{s}\) . From point \(B,\) it slides on a level surface a distance of 3.00 \(\mathrm{m}\) to point \(C,\) where it comes to rest (a) What is the coefficient of kinetic friction on the horizontal surface? (b) How much work is done on the package by friction as it slides down the circular are from \(A\) to \(B ?\)

A certain spring is found not to obey Hooke's law; it exerts a restoring force \(F_{x}(x)=-\alpha x-\beta x^{2}\) if it is stretched or compressed, where \(\alpha=60.0 \mathrm{N} / \mathrm{m}\) and \(\beta=18.0 \mathrm{N} / \mathrm{m}^{2} .\) The mass of the spring is negligible. (a) Calculate the potentinl-energy function \(U(x)\) for this spring. Let \(U=0\) when \(x=0\) (b) An object with mass 0.900 \(\mathrm{kg}\) on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 \(\mathrm{m}\) to the right (the \(+x\) -direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 \(\mathrm{m}\) to the right of the \(x=0\) equilibrium position?

Gravity in Three Dimensions. A point mass \(m_{1}\) is held in place at the origin, and another point mass \(m_{2}\) is free to move a distance \(r\) away at a point \(P\) having coordinates \(x, y,\) and \(z\) . The gravitational potential energy of these masses is found to be \(U(r)=-G m_{1} m_{2} / r,\) where \(G\) is the gravitational constant (see Exercises 7.34 and 7.35 . (a) Show that the components of the force on \(m_{2}\) due to \(m_{1}\) are $$ F_{x}=-\frac{G m_{1} m_{2} x}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} \quad F_{y}=-\frac{G m_{1} m_{2} y}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} $$ $$ F_{z}=-\frac{G m_{1} m_{2} z}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} $$ (Hint: First write \(r\) in terms of \(x, y,\) and \(z\) ) (b) Show that the magnitude of the force on \(m_{2}\) is \(F=G m_{1} m_{2} / r^{2} .\left(\text { c) Does } m_{1} \text { attract or }\right.\) repel \(m_{2} ?\) How do you know?

Two blocks with different mass are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended 1.20 \(\mathrm{m}\) , its speed is 3.00 \(\mathrm{m} / \mathrm{s}\) . If the total mass of the two blocks is \(15.0 \mathrm{kg},\) what is the mass of each block?

A \(10.0-\mathrm{kg}\) box is pulled by a horizontal wire in a circle on a rough horizontal surface for which the coefficient of kinetic friction is 0.250 . Calculate the work done by friction during one complete circular trip if the radius is (a) 2.00 \(\mathrm{m}\) and (b) 4.00 \(\mathrm{m}\) . (c) On the basis of the results you just obtained, would you say that friction is a conservative or nonconservative force? Explain.

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.