Question

A player passes a \(0.600-\mathrm{kg}\) basketball down court for a fast break. The ball leaves the player's hands with a speed of \(8.30 \mathrm{~m} / \mathrm{s}\) and slows down to \(7.10 \mathrm{~m} / \mathrm{s}\) at its highest point. Ignoring air resistance, how high above the release point is the ball when it is at its maximum height?

Short answer

The ball reaches a height of approximately 5.6 meters above the release point.

Step-by-step solution

Identify key values

The mass of the basketball is given as \(0.600\, \text{kg}\), but we don't need it for this energy problem. The initial speed of the basketball \( v_1 \) is \(8.30\, \text{m/s}\) and the final speed \( v_2 \) at the highest point is \(7.10\, \text{m/s}\). We are looking for the height \( h \) above the release point.

Apply the work-energy principle

The work-energy principle states that the change in kinetic energy is equal to the negative change in potential energy, as there is no external horizontal work done. The formula is \( \Delta KE = -\Delta PE \). Here \( KE_i - KE_f = - (PE_f - PE_i) \).

Calculate the change in kinetic energy

Kinetic energy is given by \( KE = \frac{1}{2}mv^2 \). Calculate the initial and final kinetic energies: \[ KE_i = \frac{1}{2}(0.600\, \text{kg})(8.30\, \text{m/s})^2 \]\[ KE_f = \frac{1}{2}(0.600\, \text{kg})(7.10\, \text{m/s})^2 \]

Solve for the change in potential energy

Using \( \Delta KE = KE_f - KE_i \) and knowing that \( \Delta PE = mgh \), set up the equation: \[ mgh = KE_i - KE_f \] Plug in the calculated kinetic energies and solve for \( h \):\[ h = \frac{KE_i - KE_f}{mg} \] where \( g \approx 9.81\, \text{m/s}^2 \).

Calculate the height above the release point

Substitute the known values into the equation:\[ h = \frac{(\frac{1}{2}(0.6)(8.30)^2 - \frac{1}{2}(0.6)(7.10)^2)}{(0.6)(9.81)} \]After calculations, \( h \approx 5.6 \text{ meters} \).

Key concepts

Kinetic Energy

Kinetic energy is all about the movement of objects. When an object is moving, it possesses kinetic energy. In mathematical terms, kinetic energy is given by the equation\[ KE = \frac{1}{2}mv^2 \]where \( m \) is the mass of the object and \( v \) is its velocity. The unit of kinetic energy is Joules, a standard unit of energy. When you look at an object's speed, you can directly relate it to how much kinetic energy it has.

• A faster object has more kinetic energy.
• If an object is at rest, its kinetic energy is zero.

In our basketball example, the basketball has an initial speed of \(8.30 \) meters per second when thrown and a final speed of \(7.10 \) meters per second at its highest point. We can calculate the kinetic energy at these two points to see how the energy was converted as the ball rose.

Potential Energy

Potential energy is the energy stored in an object due to its position relative to some reference point. For objects near the Earth's surface, potential energy is often related to height and is given by the formula\[ PE = mgh \]where \( m \) is mass, \( g \) is the acceleration due to gravity (approximately \( 9.81 \) meters per second squared), and \( h \) is the height above the reference point.
Potential energy increases when an object is lifted higher.

• Higher positions equals greater potential energy.
• As the basketball rises, it converts kinetic energy into potential energy.

In the basketball problem, potential energy plays a crucial role as the ball ascends to its peak height, storing energy that was previously kinetic.

Physics Problem Solving

Solving physics problems involves more than just plugging numbers into formulas. It's about understanding concepts and how they relate to each other. Here are the steps you generally follow:

• Identify what is given and what you need to find.
• Use appropriate formulas that connect what is given to what needs to be found.
• Make sure the units are consistent.
• Solve algebraically before substituting values.
• Double-check your work for accuracy.

In the case of our basketball exercise, we used the Work-Energy Principle to link kinetic energy to potential energy. Recognizing this key connection allowed us to solve for the maximum height the basketball reached in a logical and mathematical way.

Projectile Motion

Projectile motion describes the path of an object that is thrown or propelled and is subject only to acceleration due to gravity. It involves two components of motion: horizontal and vertical. The basketball thrown by the player is a classic example.

• Horizontally, the ball will move at a constant speed if we ignore forces like air resistance.
• Vertically, the ball is affected by gravity, slowing as it rises and accelerating as it falls.

In our problem, at its highest point, the basketball's vertical speed momentarily becomes zero before it starts descending. This indicates the maximum height reached, which was calculated by understanding the transfer of kinetic to potential energy.