Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 38

A cannonball of mass \(5.99 \mathrm{~kg}\) is shot from a cannon at an angle of \(50.21^{\circ}\) relative to the horizontal and with an initial speed of \(52.61 \mathrm{~m} / \mathrm{s}\). As the cannonball reaches the highest point of its trajectory, what is the gain in its potential energy relative to the point from which it was shot?

Short Answer

Expert verified
Answer: To calculate the gain in potential energy, follow these steps: 1. Calculate the vertical component of the initial velocity using the formula: \(v_{0y} = v_0 \times \sin(\theta)\). 2. Convert the angle from degrees to radians using the formula: \(\theta(rad) = \frac{\theta(deg) \times \pi}{180}\). 3. Calculate the time taken to reach the highest point using the formula: \(t = \frac{v_{0y}}{g}\). 4. Calculate the maximum height above the ground using the formula: \(h = v_{0y}t - \frac{1}{2}gt^2\). 5. Calculate the potential energy gained using the formula: \(\Delta PE = mgh\). By following these steps, you will find the gain in potential energy at the highest point of the cannonball's trajectory.

Step by step solution

01

Calculate the vertical component of the initial velocity

Using the angle of projection (50.21°) and the initial speed (52.61 m/s), we can determine the initial vertical velocity component. We use the trigonometric sine function: $$ v_{0y} = v_0 \times \sin(\theta) $$ where \(v_0\) is the initial speed (52.61 m/s), \(\theta\) is the angle (50.21°), and \(v_{0y}\) is the vertical component of the velocity.
02

Convert the angle from degrees to radians

To use the sine function, we need to convert the angle from degrees to radians: $$ \theta(rad) = \frac{\theta(deg) \times \pi}{180} $$
03

Calculate the time taken to reach the highest point

The time taken to reach the highest point can be calculated using the kinematic formula: $$ v_y = v_{0y} - gt $$ At the highest point, the vertical velocity (\(v_y\)) will be 0. So solving for \(t\), where \(g\) is the gravitational acceleration (9.81 m/s²): $$ t = \frac{v_{0y}}{g} $$
04

Calculate the maximum height above the ground

Use the time calculated above to find the maximum height reached by the cannonball: $$ h = v_{0y}t - \frac{1}{2}gt^2 $$
05

Calculate the potential energy gained

The gain in potential energy is found by multiplying the mass, gravitational acceleration, and the maximum height: $$ \Delta PE = mgh $$ Where \(m\) is the mass of the cannonball (5.99 kg), \(g\) is gravitational acceleration (9.81 m/s²), and \(h\) is the maximum height.

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!

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 truck of mass 10,212 kg moving at a speed of \(61.2 \mathrm{mph}\) has lost its brakes. Fortunately, the driver finds a runaway lane, a gravel-covered incline that uses friction to stop a truck in such a situation; see the figure. In this case, the incline makes an angle of \(\theta=40.15^{\circ}\) with the horizontal, and the gravel has a coefficient of friction of 0.634 with the tires of the truck. How far along the incline \((\Delta x)\) does the truck travel before it stops?

The energy height, \(H\), of an aircraft of mass \(m\) at altitude \(h\) and with speed \(v\) is defined as its total energy (with the zero of the potential energy taken at ground level) divided by its weight. Thus, the energy height is a quantity with units of length. a) Derive an expression for the energy height, \(H\), in terms of the quantities \(m, h\), and \(v\). b) A Boeing 747 jet with mass \(3.5 \cdot 10^{5} \mathrm{~kg}\) is cruising in level flight at \(250.0 \mathrm{~m} / \mathrm{s}\) at an altitude of \(10.0 \mathrm{~km} .\) Calculate the value of its energy height. Note: The energy height is the maximum altitude an aircraft can reach by "zooming" (pulling into a vertical climb without changing the engine thrust). This maneuver is not recommended for a 747 , however.

A particle is moving along the \(x\) -axis subject to the potential energy function \(U(x)=1 / x+x^{2}+x-1\) a) Express the force felt by the particle as a function of \(x\). b) Plot this force and the potential energy function. c) Determine the net force on the particle at the coordinate \(x=2.00 \mathrm{~m}\)

A snowboarder of mass \(70.1 \mathrm{~kg}\) (including gear and clothing), starting with a speed of \(5.1 \mathrm{~m} / \mathrm{s}\), slides down a slope at an angle \(\theta=37.1^{\circ}\) with the horizontal. The coefficient of kinetic friction is \(0.116 .\) What is the net work done on the snowboarder in the first 5.72 s of descent?

A package is dropped on a horizontal conveyor belt. The mass of the package is \(m,\) the speed of the conveyor belt is \(v\), and the coefficient of kinetic friction between the package and the belt is \(\mu_{\mathrm{k}}\) a) How long does it take for the package to stop sliding on the belt? b) What is the package's displacement during this time? c) What is the energy dissipated by friction? d) What is the total work supplied by the system?
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Physical Quantities And Units

Read Explanation

Thermodynamics

Read Explanation

Engineering Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Quantum Physics

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