Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 7: Problem 56

A ball is thrown upward with an initial velocity of 15 m/s at an angle of 60.0\(^\circ\) above the horizontal. Use energy conservation to find the ball's greatest height above the ground.

Short Answer

Expert verified
The ball reaches a height of approximately 8.60 meters.

Step by step solution

01

Understand the Problem

We need to find the greatest height reached by the ball when thrown upwards at an angle of 60 degrees with an initial speed of 15 m/s. We will use the principle of conservation of energy to find this height.
02

Define the Energy Conservation Equation

According to the conservation of mechanical energy, the initial kinetic energy (KE) of the ball will be converted into gravitational potential energy (PE) at its highest point. We have:\[\text{Initial } KE = \frac{1}{2} m v_i^2 \quad \text{and} \quad \text{Final } PE = mgh\]Where \(v_i\) is the initial vertical component of velocity, \(m\) is the mass of the ball, \(g\) is the acceleration due to gravity, and \(h\) is the unknown height.
03

Calculate the Initial Vertical Velocity Component

The vertical component of the initial velocity \(v_{i, y}\) can be calculated as:\[v_{i, y} = v_i \cdot \sin(\theta)\]Where \(v_i = 15\, \text{m/s}\) and \(\theta = 60.0^{\circ}\). Calculating gives:\[v_{i, y} = 15 \cdot \sin(60^{\circ}) = 15 \cdot \frac{\sqrt{3}}{2} \approx 12.99\, \text{m/s}\]
04

Relate Initial Kinetic and Final Potential Energy

Set the initial kinetic energy equal to the final potential energy:\[\frac{1}{2} m (v_{i, y})^2 = mgh\]Because the mass \(m\) is on both sides of the equation, it cancels out. We then have:\[\frac{1}{2} (12.99)^2 = gh\]
05

Solve for the Height

We can now solve for \(h\) using the equation:\[h = \frac{(12.99)^2}{2 \times 9.81}\]Calculating this gives:\[h = \frac{168.7401}{19.62} \approx 8.60 \text{ m}\]
06

Conclusion

The greatest height the ball reaches above the ground is approximately 8.60 meters. The calculations are based on converting all initial kinetic energy into potential energy at the highest point.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When the ball is thrown upwards, it starts with a certain amount of kinetic energy, which is determined by its initial velocity and mass. The formula for kinetic energy is:\[ KE = \frac{1}{2}mv^2 \]At the beginning, the ball has maximal kinetic energy because it is moving quickly. As an object moves, especially when it is thrown upwards, its kinetic energy decreases because its speed reduces until it brings the object to a temporary stop at its peak height.- **Initial Kinetic Energy**: This is the energy when the ball just left your hand. This is when the ball is set in motion with the initial velocity split into vertical and horizontal components.- **Conversion to Potential Energy**: Important to note is that as the ball rises, its kinetic energy converts into potential energy due to the height it gains.Understanding kinetic energy helps us track how energy is transferred and transformed during the ball's motion.
Potential Energy
Potential energy is energy stored due to an object's position or height. In our exercise, when the ball reaches its maximum height, all of its initial kinetic energy has been converted into potential energy. The formula for gravitational potential energy is:\[ PE = mgh \]- **Gravitational Potential Energy**: This type of potential energy increases as the ball rises higher in the air. Here, 'h' is the height above the ground, 'm' is the mass of the ball, and 'g' is the gravitational pull of the earth, which is approximately 9.81 m/s². - **Energy Conservation**: As energy is conserved, potential energy at the peak will equal the initial kinetic energy (when air resistance is neglected).- **Maximum Height**: At the maximum height, kinetic energy is zero, and potential energy is at its peak. Knowing potential energy helps us calculate how high the ball goes by determining how much energy is converted from kinetic to potential energy.
Vertical Velocity
Vertical velocity is the component of the initial velocity that influences the ball’s height. It is affected by the angle at which an object is thrown. For the ball thrown at an angle, the initial velocity has two components—horizontal and vertical.- **Vertical Component Calculation**: The formula to find the vertical component of the initial velocity is: \[ v_{i,y} = v_i \cdot \sin(\theta) \] Here, \(v_i\) is the initial velocity of 15 m/s and \(\theta\) is the angle of 60 degrees.- **Effect on Motion**: The vertical component influences how high the object will go. The larger this component, the higher the ball will reach.- **Stopping Point**: The vertical velocity decreases as the ball ascends, eventually reaching zero at the ball's peak height.Understanding vertical velocity is crucial because it allows us to calculate the maximum height using energy conservation. It directly affects how kinetic energy is distributed as the ball ascends.

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 10.0-kg microwave oven is pushed 6.00 m up the sloping surface of a loading ramp inclined at an angle of 36.9\(^\circ\) above the horizontal, by a constant force \(\overrightarrow{F}\) with a magnitude 110 N and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.250. (a) What is the work done on the oven by the force \(\overrightarrow{F}\)? (b) What is the work done on the oven by the friction force? (c) Compute the increase in potential energy for the oven. (d) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven's kinetic energy. (e) Use \(\sum \overrightarrow{F} = m\overrightarrow{a}\) to calculate the oven's acceleration. Assuming that the oven is initially at rest, use the acceleration to calculate the oven's speed after the oven has traveled 6.00 m. From this, compute the increase in the oven's kinetic energy, and compare it to your answerfor part (d).

A small block with mass 0.0400 kg slides in a vertical circle of radius \(R =\) 0.500 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point \(A\), the normal force exerted on the block by the track has magnitude 3.95 N. In this same revolution, when the block reaches the top of its path, point \(B\), the normal force exerted on the block has magnitude 0.680 N. How much work is done on the block by friction during the motion of the block from point \(A\) to point \(B\)?

A conservative force \(\overrightarrow{F}\) is in the \(+x\)-direction and has magnitude \(F(x) = a/(x + x_0)^2\), where \(\alpha = 0.800\) N \(\cdot\) m\(^2\) and \(x_0 = 0.200\) m. (a) What is the potential-energy function \(U(x)\) for this force? Let \(U(x) \rightarrow 0\) as \(x \rightarrow \infty\). (b) An object with mass \(m = 0.500\) kg is released from rest at \(x = 0\) and moves in the \(+x\)-direction. If \(\overrightarrow{F}\) is the only force acting on the object, what is the object's speed when it reaches \(x = 0.400\) m?

The Great Sandini is a 60-kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1100 N/m that he will compress with a force of 4400 N. The inside of the gun barrel is coated with Teflon, so the average friction force will be only 40 N during the 4.0 m he moves in the barrel. At what speed will he emerge from the end of the barrel, 2.5 m above his initial rest position?

A crate of mass \(M\) starts from rest at the top of a frictionless ramp inclined at an angle \(\alpha\) above the horizontal. Find its speed at the bottom of the ramp, a distance \(d\) from where it started. Do this in two ways: Take the level at which the potential energy is zero to be (a) at the bottom of the ramp with \(y\) positive upward, and (b) at the top of the ramp with y positive upward. (c) Why didn't the normal force enter into your solution?
See all solutions

Recommended explanations on Physics Textbooks

Quantum Physics

Read Explanation

Scientific Method Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Fundamentals of Physics

Read Explanation

Nuclear Physics

Read Explanation

Thermodynamics

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.

Sign-up for free