Question

A stone weighing \(1 / 2 \mathrm{lb}\) is thrown upward from an initial height of \(5 \mathrm{ft}\) with an initial speed of 32 \(\mathrm{ft} / \mathrm{s}\). Air resistance is proportional to speed, with \(k=1 / 128 \mathrm{lb}-\mathrm{s} / \mathrm{ft}\). Find the maximum height attained by the stone.

Short answer

Answer: The maximum height attained by the stone is approximately 50.64 ft.

Step-by-step solution

Determine the differential equation for the stone's motion

Since we are given that air resistance is proportional to speed, we can write the equation of motion for the stone as: \(m \frac{d^2x}{dt^2} = -mg - kv \Rightarrow \frac{1}{2} \frac{d^2x}{dt^2} = -\frac{1}{2}g - \frac{1}{128}v\) where \(m = 1/2\,\mathrm{lb}\) is the stone's mass, \(x\) is the height, \(v = \frac{dx}{dt}\) is the velocity, \(g = 32\, \mathrm{ft/s^2}\) is the acceleration due to gravity, and \(k = 1/128 \,\mathrm{lb\ s /ft}\) is the air resistance coefficient.

Solve the differential equation for the velocity

Rewriting the equation in terms of velocity, we get: \(\frac{dv}{dt} = -32 - \frac{1}{64}v\) Separating the variables and integrating, we obtain: \(\int \frac{1}{32 + \frac{1}{64}v} dv = -\int dt\) Using substitution, let \(u = 32 + \frac{1}{64}v \Rightarrow dv = 64du\) and integrate: \(\int\frac{64}{u}du = -\int dt\) \(64(\ln|u|+C_1) = -t+C_2\) We now solve for the constants \(C_1\) and \(C_2\) using the initial condition \(v(0) = 32\, \mathrm{ft/s}\): Substituting \(t=0, v=32\), we find \(C_2 = 64 (\ln(32)+C_1)\). Now we solve for velocity: \(v(t) = -64 + \frac{2048}{32+e^{-\frac{t}{64}}}\)

Calculate the height function

To find the height function \(x(t)\), we integrate the velocity function: \(x(t) = \int v(t) dt\) Integrating and substituting back our values, we get: \(x(t) = -64t + 64^2 \ln(32+e^{-\frac{t}{64}}) + C\) To solve for the arbitrary constant \(C\), we use the initial condition \(x(0) = 5\, \mathrm{ft}\): \(x(0) = -64(0) + 64^2 \ln(32+e^{0}) + C = 5\) Thus, \(C = 5-64^2 \ln(33)\), and we find: \(x(t) = -64t + 64^2 \ln(32+e^{-\frac{t}{64}}) - 64^2 \ln(33) + 5\)

Determine the maximum height attained by the stone

The maximum height is reached when the stone's velocity is zero. Thus, we need to find the time at which \(\displaystyle v(t)=0\): \(0 = -64 + \frac{2048}{32+e^{-\frac{t}{64}}}\) Solving for time \(t\), we get \(\displaystyle t\approx 78.29\, \mathrm{s}\). Now, we substitute this value into the height function: \(x(78.29) \approx -64(78.29) + 64^2 \ln(32+e^{-\frac{78.29}{64}}) - 64^2 \ln(33) + 5 \approx 50.64\, \mathrm{ft}\) So, the maximum height attained by the stone is \(\boxed{50.64\, \mathrm{ft}}\).

Key concepts

Motion with Air Resistance

Understanding the motion of objects through the air is vastly improved when we consider the effects of air resistance. In our everyday experience, objects that we throw or drop are slowed down by the air, which is why a feather falls more slowly than a stone. When an object moves through the air, air resistance is often proportional to its speed; that is, the faster the object moves, the greater the resisting force exerted by the air.

In the context of our problem, we're dealing with a stone that not only experiences the force of gravity but also a retarding force due to air resistance. The key point to remember is that air resistance affects the acceleration of the stone, which directly impacts its velocity over time and hence the maximum height it reaches. The air resistance is given by the relationship \( kv \), where \( k \) represents the constant of proportionality and \( v \) its velocity.

Ordinary Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In physics and engineering, differential equations are indispensable for modeling how systems evolve over time. An ordinary differential equation (ODE) is a type of differential equation that contains one or more functions of one independent variable and its derivatives.

The problem at hand involves a second-order ODE, since we are dealing with the second derivative of the stone’s height with respect to time. The equation of motion is given by \( m \frac{d^2x}{dt^2} = -mg - kv \), signifying that the motion of the stone is affected both by gravity and air resistance. For physical systems like this, ODEs are solved to predict behavior such as the stone’s position and velocity at any time \( t \).

Initial Value Problem

An initial value problem is a type of differential equation along with a set of initial conditions. Solving the differential equation typically involves finding a function that not only satisfies the equation but also meets the specified initial conditions. In the case of our problem, we're given the initial conditions that the stone starts from a height of 5 feet with an upward speed of 32 feet per second.

These initial conditions are crucial because they allow us to uniquely determine the constants of integration that arise when solving the ODE, ensuring that the solution is specific to the problem we're trying to solve. Without initial conditions, we would have an infinite number of potential solutions, which wouldn't give us the precise information we need about the stone's motion.

Exponential Decay

Exponential decay describes a process where the amount of something decreases at a rate proportional to its current value. This concept is widely used across disciplines, including physics, chemistry, and finance. When we solve the differential equation related to our problem, we encounter an exponential function, which arises from integrating the velocity function that contains an exponential term.

The velocity of the stone, adjusted for air resistance, includes an exponential decay component, which affects how quickly it slows down as it ascends. It's the interplay between the force of gravity and the resistance of air that creates this exponential decrease in velocity. Determining the time when the velocity is zero gives us the information we need to calculate the maximum height the stone attains.