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Chapter 5: Problem 14

Use bisection to determine the drag coefficient needed so that an 80 -kg parachutist has a velocity of \(36 \mathrm{m} / \mathrm{s}\) after \(4 \mathrm{s}\) of free fall. Note: The acceleration of gravity is \(9.81 \mathrm{m} / \mathrm{s}^{2}\). Start with initial guesses of \(x_{1}=0.1\) and \(x_{u}=0.2\) and iterate until the approximate relative error falls below \(2 \%\).

Short Answer

Expert verified
The drag coefficient needed for an 80-kg parachutist to have a velocity of 36 m/s after 4 seconds of free fall can be determined using the bisection method on the function \(f(c) = v_d - v(t)\), based on the quadratic drag force equation \(v(t) = \dfrac{gm}{c}(1 - e^{-\dfrac{c t}{m}})\). With initial guesses \(x_{l} = 0.1\) and \(x_{u} = 0.2\), iterate the bisection method until the approximate relative error falls below 2%. The resulting value of the drag coefficient is the midpoint value \(x_{r}\) of the final interval.

Step by step solution

01

Define the function and its parameters

Define the function \(v(t) = \dfrac{gm}{c}(1 - e^{-\dfrac{c t}{m}})\) with the given values: \(g = 9.81 \mathrm{m} / \mathrm{s}^{2}\), \(m = 80 \mathrm{kg}\), \(t = 4 \mathrm{s}\), and the desired velocity \(v_d = 36 \mathrm{m} / \mathrm{s}\). For the bisection method, we will use the signed difference between the desired velocity and the actual velocity given by the function: \(f(c) = v_d - v(t)\).
02

Initialize bisection method

Start with initial guesses \(x_{l} = 0.1\) and \(x_{u} = 0.2\). Calculate the signed difference for both guesses, \(f(x_{l})\) and \(f(x_{u})\).
03

Perform bisection iterations

Iterate using the following steps until the approximate relative error is below 2%: 1. Calculate the midpoint value \(x_{r} = \dfrac{x_{l} + x_{u}}{2}\). 2. Calculate the signed difference at the midpoint, \(f(x_{r})\). 3. Check the product of \(f(x_{l}) \cdot f(x_{r})\). - If the product is negative, set \(x_{u} = x_{r}\). - If the product is positive, set \(x_{l} = x_{r}\). - If the product is zero, the solution is \(x_{r}\). 4. Calculate the approximate relative error: \(\dfrac{\left|x_{u} - x_{l}\right|}{x_{u}} \times 100\%\). 5. Repeat from step 1 until approximate relative error is below 2%.
04

Determine the drag coefficient

After the bisection iterations, the appropriate value of the drag coefficient is the midpoint value \(x_{r}\). This is the drag coefficient, which when put into the equation, results in a velocity of 36 m/s after 4 seconds of free fall.

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Key Concepts

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

Drag Coefficient
The drag coefficient, often symbolized as \(c\), is a crucial factor in fluid dynamics. It determines how resistant an object is to motion through a fluid, such as air. For a parachutist, the drag coefficient affects how quickly they can reach a stable velocity while falling freely. Manipulating this coefficient results in controlling the descent speed, primarily how it counteracts gravitational forces. The lower the drag coefficient, the less air resistance experienced, leading to faster speeds, while a higher drag coefficient indicates more resistance and slower speeds. Understanding this parameter is essential for calculating realistic and safe velocities for objects in free fall scenarios.
Parachutist Free Fall
In physics, parachutist free fall refers to the phase of descent when a parachutist jumps from an aircraft and gravity is the predominant force acting on them. During this time, air resistance progressively increases until it balances with the gravitational force, leading to a terminal velocity. The velocity of a parachutist during free fall depends on several factors: mass, gravitational pull (which, according to the exercise, is \(9.81 \mathrm{m} / \mathrm{s}^{2}\)), and the drag coefficient. Understanding this phase is vital, as it allows one to calculate the expected descent speed before deploying a parachute. It ensures safety and precision in various jumping conditions.
Numerical Methods
Numerical methods, such as the bisection method used in this exercise, are mathematical techniques aimed at finding approximate solutions to complex mathematical problems. They are especially useful when an exact solution is impossible or impractical to obtain analytically. The bisection method involves narrowing down the range (or interval) where a function changes sign, i.e., crosses zero. The method requires:
  • Choosing initial guesses that bracket the root.
  • Calculating the midpoint.
  • Identifying the subinterval containing the root by checking the sign.
  • Repeating the process until the error becomes acceptably small.
For this particular problem, utilizing the bisection method allows us to effectively determine the drag coefficient by targeting the desired velocity value.
Velocity Calculation
Calculating velocity in the context of parachutist free fall involves a blend of gravitational effects and resistance caused by drag. The velocity at any moment can be represented by the equation: \[ v(t) = \dfrac{gm}{c}(1 - e^{-\dfrac{ct}{m}}) \]where:
  • \( g \) is gravity, \( 9.81 \mathrm{m} / \mathrm{s}^2 \).
  • \( m \) is the mass of the parachutist.
  • \( c \) is the drag coefficient.
  • \( t \) is time.
In our problem, we aim to achieve a velocity of \(36 \mathrm{m} / \mathrm{s}\) after \(4\) seconds. By accurately finding the drag coefficient using our equation and numerical methods, we adjust how quickly the parachutist reaches the desired speed. This process combines physics and mathematical strategies to ensure controlled and predictable descent.

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Most popular questions from this chapter

You buy a \(\$ 25,000\) piece of equipment for nothing down and \(\$ 5,500\) per year for 6 years. What interest rate are you paying? The formula relating present worth \(P,\) annual payments \(A,\) number of years \(n,\) and interest rate \(i\) is \\[ A=P \frac{i(1+i)^{n}}{(1+i)^{n}-1} \\]

Determine the real root of \(x^{3.5}=80:\) (a) analytically, and (b) with the false-position method to within \(\varepsilon_{s}=2.5 \%\). Use initial guesses of 2.0 and 5.0.

Figure \(\mathrm{P} 5.20\) a shows a uniform beam subject to a linearly increasing distributed load. The equation for the resulting elastic curve is (see Fig. P5.20 \(b\) ) $$y=\frac{w_{0}}{120 E I L}\left(-x^{5}+2 L^{2} x^{3}-L^{4} x\right)$$ Use bisection to determine the point of maximum deflection (that is, the value of \(x\) where \(d y / d x=0\) ). Then substitute this value into Eq. (P5.20) to determine the value of the maximum deflection. Use the following parameter values in your computation: \(L=600 \mathrm{cm}\) \(E=50,000 \mathrm{kN} / \mathrm{cm}^{2}, I=30,000 \mathrm{cm}^{4},\) and \(w_{0}=2.5 \mathrm{kN} / \mathrm{cm}\)

Determine the real root of \(f(x)=-26+85 x-91 x^{2}+\) \(44 x^{3}-8 x^{4}+x^{5}:\) (a) Graphically. (b) Using bisection to determine the root to \(\varepsilon_{s}=10 \%\). Employ initial guesses of \(x_{1}=0.5\) and \(x_{u}=1.0\) (c) Perform the same computation as in (b) but use the false-position method and \(\varepsilon_{s}=0.2 \%\).

The saturation concentration of dissolved oxygen in freshwater can be calculated with the equation (APHA, 1992 ) $$\begin{aligned} \ln o_{s f}=&-139.34411+\frac{1.575701 \times 10^{5}}{T_{a}}-\frac{6.642308 \times 10^{7}}{T_{a}^{2}} \\ &+\frac{1.243800 \times 10^{10}}{T_{a}^{3}}-\frac{8.621949 \times 10^{11}}{T_{a}^{4}} \end{aligned}$$ where \(o_{s f}=\) the saturation concentration of dissolved oxygen in freshwater at 1 atm \((\mathrm{mg} / \mathrm{L})\) and \(T_{a}=\) absolute temperature \((\mathrm{K})\) Remember that \(T_{a}=T+273.15,\) where \(T=\) temperature \(\left(^{\circ} \mathrm{C}\right)\) According to this equation, saturation decreases with increasing temperature. For typical natural waters in temperate climates, the equation can be used to determine that oxygen concentration ranges from \(14.621 \mathrm{mg} / \mathrm{L}\) at \(0^{\circ} \mathrm{C}\) to \(6.413 \mathrm{mg} / \mathrm{L}\) at \(40^{\circ} \mathrm{C}\). Given a value of oxygen concentration, this formula and the bisection method can be used to solve for temperature in \(^{\circ} \mathrm{C}\) (a) If the initial guesses are set as 0 and \(40^{\circ} \mathrm{C}\), how many bisection iterations would be required to determine temperature to an absolute error of \(0.05^{\circ} \mathrm{C} ?\) (b) Develop and test a bisection program to determine \(T\) as a function of a given oxygen concentration to a prespecified absolute error as in (a). Given initial guesses of 0 and \(40^{\circ} \mathrm{C}\), test your program for an absolute error \(=0.05^{\circ} \mathrm{C}\) and the following cases: \(o_{s f}=8,10\) and \(12 \mathrm{mg} / \mathrm{L}\). Check your results.
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