Question

An object is dropped from altitude \(y_{0}\). (a) Assume that the drag force is proportional to velocity, with drag coefficient \(k\). Obtain an implicit solution relating velocity and altitude. (b) If the terminal velocity is known to be \(-120 \mathrm{mph}\) and the impact velocity was \(-90 \mathrm{mph}\), what was the initial altitude \(y_{0}\) ?

Short answer

Question: An object falls vertically under the influence of gravity and air resistance. If the drag force is proportional to its velocity, find the implicit solution relating velocity and altitude. Given the terminal velocity is -120 mph and the impact velocity is -90 mph, find the initial altitude (y_0) from which the object was dropped. Answer: To find the implicit solution relating velocity and altitude, first write the differential equation for the object's motion: \(v\frac{dv}{dy} = g - (\frac{k}{m})v\). Integrate to get \(-\frac{m}{k} \ln|g - (\frac{k}{m})v| \Big|_{v_0}^v = y - y_0\). To find the initial altitude, plug in the terminal and impact velocities and solve for \(y_0\): \(y_0 = \frac{m}{k}(\ln|g - (\frac{k}{m})(-90 \mathrm{mph})| - \ln|g - (\frac{k}{m})(-120 \mathrm{mph})|) + y_i\). Calculate the value of y_0 by inserting the corresponding values for g, k, and m, and convert the velocities to the same unit if needed. This will give you the initial altitude, y_0.

Step-by-step solution

Write down the applicable equations

Consider an object falling vertically under the influence of gravity and air resistance. The forces acting on it are gravity and drag force. Assuming downward direction is positive, we can write the Newton's second law equation as follows: \(ma = mg - kv\) Where \(a\) is the object's acceleration, \(m\) is the object's mass, \(g\) is the acceleration due to gravity, \(k\) is the drag coefficient, and \(v\) is the object's downward velocity.

Solve the differential equation for velocity

To find the velocity as a function of altitude, we need to solve the following differential equation: \(m\frac{dv}{dy} = mg - kv\) To make it easier to solve, change the variable, so the equation is in terms of \(y\). To achieve this, we start by rearranging the equation: \(\frac{dv}{dt} = g - (\frac{k}{m})v\) Now, replace \(\frac{dv}{dt}\) by \(\frac{dv}{dy} \cdot \frac{dy}{dt}\) and note that, \(\frac{dy}{dt} = v\), so we get: \(v\frac{dv}{dy} = g - (\frac{k}{m})v\) Now we can solve the first order linear differential equation: \(\frac{dv}{g - (\frac{k}{m})v} = dy\)

Integrate both sides of the equation

Integrate both sides of the equation to obtain an implicit solution relating velocity and altitude: \(\int_{v_0}^v \frac{dv'}{g - (\frac{k}{m})v'} = \int_{y_0}^y dy\) This can be solved using a substitution method. Let \(u = g - (\frac{k}{m})v'\) and \(du = -(\frac{k}{m})dv'\) : \(\int_{v_0}^v \frac{-\frac{m}{k} du}{u} = \int_{y_0}^y dy\) Integrating both sides, we get: \(-\frac{m}{k} \ln|g - (\frac{k}{m})v| \Big|_{v_0}^v = y - y_0\) This is the desired implicit solution relating velocity (\(v\)) and altitude (\(y\)).

Compute initial altitude using terminal and impact velocities

To find the initial altitude \(y_0\), we will use the given terminal velocity \(v_t = -120 \mathrm{mph}\), and the impact velocity \(v_i = -90 \mathrm{mph}\). Plug the terminal velocity into the implicit solution and solve for \(y_t\): \(-\frac{m}{k} \ln|g - (\frac{k}{m})(-120 \mathrm{mph})| = y_t - y_0\) Now, plug the impact velocity into the implicit solution and solve for \(y_i\): \(-\frac{m}{k} \ln|g - (\frac{k}{m})(-90 \mathrm{mph})| = y_i - y_0\) Use those two equations to find the initial altitude \(y_0\). Subtract the second equation from the first: \(-\frac{m}{k} (\ln|g - (\frac{k}{m})(-120 \mathrm{mph})| - \ln|g - (\frac{k}{m})(-90 \mathrm{mph})|) = y_t - y_i\) Since the mass \(m\) and the drag coefficient \(k\) are constants, we can compute the initial altitude \(y_0\): \(y_0 = \frac{m}{k}(\ln|g - (\frac{k}{m})(-90 \mathrm{mph})| - \ln|g - (\frac{k}{m})(-120 \mathrm{mph})|) + y_i\) Find the value of y_0 by plugging in the corresponding values for g, k and m and converting the velocities to the same unit, if necessary. After calculating the value, the initial altitude \(y_0\) can be obtained.

Key concepts

Drag Force

The drag force is a resisting force that acts on an object when it moves through a fluid like air or water. It opposes the object's motion, slowing it down due to friction and differences in pressure across the object. In our problem, the drag force is assumed to be proportional to the velocity of the falling object. This means the faster the object moves, the greater the opposing force it experiences.When evaluating drag force, you'll often see it represented by the equation:\[ F_d = -kv \]where:

• \(F_d\) is the drag force
• \(k\) is the drag coefficient
• \(v\) is the velocity

This proportionality simplifies our calculations, but in real-world scenarios, drag depends on several factors:

• Shape of the object
• Density of the fluid
• Area exposed to the fluid

Understanding drag force is crucial, as it affects everything from the speeds of cars and bicycles to the fall of parachutists.

Terminal Velocity

Terminal velocity refers to the constant speed that a freely falling object eventually reaches when the drag force due to air resistance equals the gravitational force. At this point, the object stops accelerating and continues to fall at this steady velocity.In our exercise, the terminal velocity is given. This indicates the exact balance between two forces:

• The downward gravitational pull \(mg\)
• The upward drag force \(kv\)

When these forces are equal, the net force is zero, so the acceleration is zero, and velocity remains constant. For terminal velocity, the force equation simplifies to:\[ mg = kv_{t} \]where \(v_{t}\) is the terminal velocity. This condition makes determining key values, like initial altitude or time of fall, easier with known velocities. It is a fundamental concept in fluid dynamics and an important topic of study when analyzing objects moving through fluid bodies.

Newton's Second Law

Newton's Second Law is a cornerstone of physics and states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, it is represented as:\[ F = ma \]In this exercise, the law is used to relate forces acting on the falling object. The main forces to consider here are gravity and the drag force.

Gravity exerts a constant force \(mg\) (where \(m\) is mass and \(g\) is acceleration due to gravity). The drag is \(-kv\)\( (k\) is drag coefficient and \(v\) is velocity).

Together, they form the equation:\[ ma = mg - kv \]which links acceleration \(a\) with velocity through the object's mass and drag coefficient.This equation helps us model how velocity changes with time and altitude. It serves as the basis for differential equations that describe motion under resistance, which are solved using integration techniques.

Integration Methods

Integration is a mathematical technique used to solve differential equations, which often describe relationships in physics. In our context, integration helps connect velocity and altitude of a falling object.Given the differential equation:\[ \frac{dv}{dy} \cdot v = g - \left(\frac{k}{m}\right)v \]Integration allows us to obtain an expression that relates velocity \(v\) and altitude \(y\). To achieve this, we separate variables:\[ \int \frac{dv}{g - \left(\frac{k}{m}\right)v} = \int dy \]Here, substitution methods make integration simpler, enhancing manageability. Once integrated, the result provides an implicit solution showing how velocity varies with altitude:\[ -\frac{m}{k} \ln\left|g - \left(\frac{k}{m}\right)v\right| = y - y_0 \]This shows the change in altitude as the object speeds up or slows down due to drag. Each integration step not only aids in solving the equation but also offers valuable insights into the object’s dynamics in a fluid environment.