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 \(1.02 \mathrm{~m}\).

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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.

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Most popular questions from this chapter

A basketball of mass \(0.624 \mathrm{~kg}\) is shot from a vertical height of \(1.2 \mathrm{~m}\) and at a speed of \(20.0 \mathrm{~m} / \mathrm{s}\). After reaching its maximum height, the ball moves into the hoop on its downward path, at \(3.05 \mathrm{~m}\) above the ground. Using the principle of energy conservation, determine how fast the ball is moving just before it enters the hoop.

A block of mass \(5.0 \mathrm{~kg}\) slides without friction at a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) on a horizontal table surface until it strikes and sticks to a mass of \(4.0 \mathrm{~kg}\) attached to a horizontal spring (with spring constant of \(k=2000.0 \mathrm{~N} / \mathrm{m}\) ), which in turn is attached to a wall. How far is the spring compressed before the masses come to rest? a) \(0.40 \mathrm{~m}\) b) \(0.54 \mathrm{~m}\) c) \(0.30 \mathrm{~m}\) d) \(0.020 \mathrm{~m}\) e) \(0.67 \mathrm{~m}\)

A large air-filled 0.100 -kg plastic ball is thrown up into the air with an initial speed of \(10.0 \mathrm{~m} / \mathrm{s}\). At a height of \(3.00 \mathrm{~m}\) the ball's speed is \(3.00 \mathrm{~m} / \mathrm{s}\). What fraction of its original energy has been lost to air friction?

A 1.00 -kg block is pushed up and down a rough plank of length \(L=2.00 \mathrm{~m},\) inclined at \(30.0^{\circ}\) above the horizontal. From the bottom, it is pushed a distance \(L / 2\) up the plank, then pushed back down a distance \(L / 4,\) and finally pushed back up the plank until it reaches the top end. If the coefficient of kinetic friction between the block and plank is \(0.300,\) determine the work done by the block against friction.

An arrow is placed on a bow, the bowstring is pulled back, and the arrow is shot straight up into the air; the arrow then comes back down and sticks into the ground. Describe all of the changes in work and energy that occur.

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