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Chapter 9: Problem 6

A nonspinning ball having a mass of 3 oz. is thrown vertically upward with a velocity of 100 mph and has zero velocity at a height 250 ft above the release point. Assume that the air drag on the ball is constant and find this constant "average" air drag. Neglect the buoyant force of air on the ball.

Short Answer

Expert verified
The constant 'average' air drag force on the ball is calculated to be \(F_{drag}\) (in lb-f).

Step by step solution

01

Convert units

First, we must convert the given measures to a unified system of units. Since we are given the mass in ounces and height in feet, the Imperial system is convenient. Convert 3 ounces to pounds (since 1 pound = 16 ounces) and the speed of 100 mph to feet per second (since 1 mph = 1.47 feet/sec).
02

Calculate Initial Energy

The initial energy of the ball is the sum of the kinetic and potential energy. The equation for kinetic energy is \(0.5 * mass * velocity^2\) and for potential energy is \( mass * gravity * height\). Plug in the appropriate values remembering that the initial height is 0; hence, the potential energy is also 0.
03

Calculate Final Energy

At the peak of its trajectory, the ball's velocity is 0, which means its kinetic energy is also 0. The only form of energy it has at this point is potential energy, which can be calculated using the formula \(mass * gravity * height\). Plug in the height of 250 feet.
04

Calculate Work Done

By the conservation of energy principle, the work done by the drag is the difference between the initial and final energy.
05

Calculate Average Air Drag

The work done by the drag force equals the force times the distance, here, the distance is 250 feet. So, solving for drag force, it equals the work done divided by the distance.

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Key Concepts

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

Air Drag
When objects move through the air, they experience a force known as air drag. This force acts opposite to the motion, slowing the object down. Air drag is influenced by factors such as the object's speed, the shape and size of the object, and the density of the air.
  • As speed increases, air drag generally increases.
  • Smooth and streamlined shapes experience less air drag.
  • Higher air density leads to greater air drag.
For the nonspinning ball in our exercise, the air drag is considered constant. By calculating the initial and final energies of the ball, we can determine the work done by air drag over the 250-foot ascent. This work is the energy lost due to the drag force. Once we know the work, we can use it to find the average air drag force by dividing the total work by the distance traveled.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It depends on both the mass and the velocity of the object.
The formula for calculating kinetic energy (KE) is: \[ KE = \frac{1}{2} \times \text{mass} \times \text{velocity}^2 \] In our exercise, the ball starts with an initial velocity of 100 mph, converted into feet per second, and has a mass converted from ounces to pounds.
  • Initial kinetic energy is calculated with all quantities converted to the same unit system.
  • At the peak, when the velocity is zero, kinetic energy becomes zero.
Understanding how kinetic energy changes helps in determining how much energy is dissipated as work done by air drag.
Potential Energy
Potential energy is the energy stored in an object due to its position, often related to its height above the ground. This type of energy can be calculated using the formula: \[ PE = \text{mass} \times \text{gravity} \times \text{height} \]
Gravity is a constant approximated as 32.2 feet per second squared in the Imperial system. At the beginning of the problem, the ball's potential energy is zero because its initial height is zero.
As the ball rises to 250 feet, its potential energy increases while its kinetic energy decreases.
  • Higher the object, more potential energy it possesses.
  • Transition from kinetic to potential energy occurs as the ball ascends.
This shift from kinetic to potential energy is crucial in understanding the conservation energy principle.
Conservation of Energy
The principle of conservation of energy states that total energy in a closed system remains constant. This means the energy at the start of the system must equal the energy at the end. In our problem, this concept helps us understand how various energy forms get converted.
  • Initial energy (kinetic + potential) equals the final energy (minus work done by drag).
  • Energy transitions from kinetic at release, to potential at the peak, not losing form but shifting location.
This principle allows for calculating air drag, as energy lost to drag creates a difference between initial and final energy. By understanding this conservation, students can apply it to similar physics problems, emphasizing its universal application across different scenarios.

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Most popular questions from this chapter

For small Reynolds number flows, the drag coefficient of an object is given by a constant divided by the Reynolds number (see Table 9.4 . Thus, as the Reynolds number tends to zero, the drag coefficient becomes infinitely large. Does this mean that for small velocities (hence, small Reynolds numbers) the drag is very large? Explain.

A \(1.2-1 b\) kite with an area of \(6 \mathrm{ft}^{2}\) flies in a \(20-\mathrm{ft} / \mathrm{s}\) wind such that the weightless string makes an angle of \(55^{\circ}\) relative to the horizontal. If the pull on the string is 1.5 lb, determine the lift and drag coefficients based on the kite area.

For many years, hitters have claimed that some baseball pitchers have the ability to actually throw a rising fastball. Assuming that a top major leaguer pitcher can throw a 95 -mph pitch and impart an 1800 -rpm spin to the ball, is it possible for the ball to actually rise? Assume the baseball diameter is 2.9 in. and its weight is 5.25 oz.

An automobile engine has a maximum power output of \(70 \mathrm{hp},\) which occurs at an engine speed of \(2200 \mathrm{rpm} . \mathrm{A} 10 \%\) power loss occurs through the transmission and differential. The rear wheels have a radius of 15.0 in. and the automobile is to have a maximum speed of 75 mph along a level road. At this speed, the power absorbed by the tires because of their continuous deformation is 27 hp. Find the maximum permissible drag coefficient for the automobile at this speed. The car's frontal area is \(24.0 \mathrm{ft}^{2}\)

A laminar boundary layer velocity profile is approximated by \(u / U=[2-(y / \delta)](y / \delta)\) for \(y \leq \delta,\) and \(u=U\) for \(y>\delta\) (a) Show that this parabolic profile satisfies the appropriate boundary conditions. (b) Use the momentum integral equation to determine the boundary layer thickness, \(\delta=\delta(x)\). Compare the result with the exact Blasius solution.
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