/*! 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 69 A high jumper approaches the bar... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A high jumper approaches the bar at \(9.0 \mathrm{~m} / \mathrm{s}\). What is the highest altitude the jumper can reach, if he does not use any additional push off the ground and is moving at \(7.0 \mathrm{~m} / \mathrm{s}\) as he goes over the bar?

Short Answer

Expert verified
Based on the conservation of energy principle, the highest altitude the high jumper can reach is approximately 1.02 meters.

Step by step solution

01

Write down the energy conservation equation

In this step, we need to write the energy conservation equation \(K_f + U_f = K_i + U_i\) as \(K_i - K_f = U_f - U_i\).
02

Calculate the initial and final kinetic energy

Using the given initial velocity (\(v_i = 9.0 \mathrm{~m} / \mathrm{s}\)) and final velocity (\(v_f = 7.0 \mathrm{~m} / \mathrm{s}\)), we will now calculate the initial and final kinetic energy using the formula \(K=\frac{1}{2}mv^2\). Due to conservation of mass, we can divide by mass \(m\) in the calculations, which cancels it out. \(K_i = \frac{1}{2}m v_i^2 = \frac{1}{2}m(9.0)^2\) \(K_f = \frac{1}{2}m v_f^2 = \frac{1}{2}m(7.0)^2\)
03

Substitute the calculated values in the energy conservation equation

Now, we substitute the values of \(K_i\) and \(K_f\) calculated in Step 2 and the initial potential energy into the energy conservation equation. Since \(U_i = 0\), we only need to consider the final potential energy, \(U_f = mgh\): \(\frac{1}{2}m(9.0)^2 - \frac{1}{2}m(7.0)^2 = mgh\)
04

Solve for the highest altitude (h)

In this step, we will solve for \(h\). We can first cancel out the mass \(m\) from both sides of the equation: \((9.0)^2 - (7.0)^2 = 2gh\) Now, we can solve for \(h\): \(h = \frac{(9.0)^2 - (7.0)^2}{2g}\) Using the value of acceleration due to gravity, \(g = 9.81 \mathrm{~m} / \mathrm{s^2}\), we can calculate the highest altitude: \(h = \frac{(9.0)^2 - (7.0)^2}{2(9.81)} \approx 1.02 \mathrm{~m}\) The highest altitude the jumper can reach is approximately \(1.02 \mathrm{~m}\).

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 that an object possesses due to its motion. It's an important concept in physics because it quantifies the amount of work that a moving object can perform due to its velocity. The formula for kinetic energy (\textbf{KE}) is given by \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.

In the context of our high jumper problem, we can see kinetic energy in action as the jumper moves towards the bar. As the velocity of the jumper changes, so does his kinetic energy. This is crucial, because, upon taking off from the ground, some of the kinetic energy is converted to potential energy, allowing the jumper to reach a certain height.
Potential Energy
Potential energy is the energy stored in an object due to its position or arrangement. Gravitational potential energy, for instance, is energy that an object possesses because of its position in a gravitational field. The formula for gravitational potential energy (\textbf{PE}) is given by \( PE = mgh \), where \( m \) is the mass, \( g \) is the acceleration due to gravity (\( 9.81 \text{m/s}^2 \) on Earth), and \( h \) is the height above the reference point.

For our high jumper, the highest altitude he can reach is determined by when all the kinetic energy, that isn’t lost to other forces, is converted to potential energy. The jumper's initial potential energy at ground level is zero which increases as he ascends.
Conservation of Mechanical Energy
The principle of conservation of mechanical energy states that the total mechanical energy of an isolated system remains constant as long as only conservative forces, like gravity, are acting on it. Mechanical energy is the sum of kinetic and potential energy. This principle can be written as \( KE_{i} + PE_{i} = KE_{f} + PE_{f} \), with ‘i’ indicating initial and ‘f’ indicating final states. Non-conservative forces, like friction or air resistance, can cause this energy to change, but they are ignored in ideal scenarios.

In the given exercise, by setting the initial potential energy on the ground to zero and considering no other forces but gravity, we applied energy conservation to solve for the maximum height the jumper could reach. The equation \( K_f + U_f = K_i + U_i \) captures the energy transformation from kinetic to potential, allowing us to predict how high the jumper will go.
Kinematics in Physics
Kinematics is the branch of mechanics that describes the motion of points, objects, and systems without considering the forces that cause them to move. It focuses on displacement, velocity, acceleration, and time. These concepts are fundamental when analyzing the motion of objects, as they help understand trajectories and motion patterns without delving into what actually propels or alters that motion.

Relating to our jumper, kinematics would not only look at the jumper's speeds at various points but could extend to the trajectory he takes throughout his jump. Even though our problem didn’t require a deep dive into kinematics, understanding how motion is described in physics provides a foundation for more complex problems where kinematics and energy conservation intertwine.

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 1.00 -kg block compresses a spring for which \(k=\) 100. \(\mathrm{N} / \mathrm{m}\) by \(20.0 \mathrm{~cm}\) and is then released to move across a horizontal, frictionless table, where it hits and compresses another spring, for which \(k=50.0 \mathrm{~N} / \mathrm{m}\). Determine a) the total mechanical energy of the system, b) the speed of the mass while moving freely between springs, and c) the maximum compression of the second spring.

A 1.00 -kg block is resting against a light, compressed spring at the bottom of a rough plane inclined at an angle of \(30.0^{\circ}\); the coefficient of kinetic friction between block and plane is \(\mu_{\mathrm{k}}=0.100 .\) Suppose the spring is compressed \(10.0 \mathrm{~cm}\) from its equilibrium length. The spring is then released, and the block separates from the spring and slides up the incline a distance of only \(2.00 \mathrm{~cm}\) beyond the spring's normal length before it stops. Determine a) the change in total mechanical energy of the system and b) the spring constant \(k\).

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.

Calculate the force \(F(y)\) associated with each of the following potential energies: a) \(U=a y^{3}-b y^{2}\) b) \(U=U_{0} \sin (c y)\)

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

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.