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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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Most popular questions from this chapter

A model is placed in an airflow at standard conditions with a given velocity and then placed in water flow at standard conditions with the same velocity. If the drag coefficients are the same between these two cases, how do the drag forces compare between the two fluids?

As is discussed in Section \(9.3,\) the drag on a rough golf ball may be less than that on an equal-sized smooth ball. Does it follow that a 10 -m-diameter spherical water tank resting on a \(20-\mathrm{m}\) -tall support should have a rough surface so as to reduce the moment needed at the base of the support when a wind blows? Explain.

Two different fluids flow over two identical flat plates with the same laminar free-stream velocity. Both fluids have the same viscosity, but one is twice as dense as the other. What is the relationship between the drag forces for these two plates?

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.

A laminar boundary layer velocity profile is approximated by \(u / U=2(y / \delta)-2(y / \delta)^{3}+(y / \delta)^{4}\) for \(y \leq \delta,\) and \(u=U\) for \(y >\delta .\) (a) Show that this 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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