Problem 24 - Differential Equations and Linear Algebra

Question

An object is released from rest at a height of 100 meters above the ground. Neglecting frictional forces, the subsequent motion is governed by the initial-value problem $$ \frac{d^{2} y}{d t^{2}}=g, \quad y(0)=0, \quad \frac{d y}{d t}(0)=0 $$where \(y(t)\) denotes the displacement of the object from its initial position at time \(t\). Solve this initial-value problem and use your solution to determine the time when the object hits the ground.

Step-by-Step Solution

Answer

The displacement equation of the object is given by \(y(t) = \frac{1}{2}gt^2\) with \(g=9.81 m/s^2\). To find when the object hits the ground, set \(y(t) = -100\) and solve for \(t\): \[-100 = \frac{1}{2}(9.81)t^2\] which yields \(t \approx 4.52 s\). The object will hit the ground after approximately \(4.52\) seconds.

Step 1: Integrate the governing equation once

By integrating the given equation, \(\frac{d^2y}{dt^2} = g\), with respect to time once, we obtain: \[\frac{dy}{dt} = gt + C_1\] where \(C_1\) is the constant of integration.

Step 2: Apply the initial condition, \(\frac{dy}{dt}(0) = 0\)

Using the given initial condition, \(\frac{dy}{dt}(0) = 0\), we can now determine the value of \(C_1\): \[0 = g(0) + C_1\] \[C_1 = 0\] Therefore, we now have: \[\frac{dy}{dt} = gt\]

Step 3: Integrate the new equation again

Now, let's integrate the equation \(\frac{dy}{dt} = gt\) once more with respect to time: \[y(t) = \int (gt) dt\] \[y(t) = \frac{1}{2}gt^2 + C_2\] where \(C_2\) is another constant of integration.

Step 4: Apply the initial condition, \(y(0) = 0\)

Using the given initial condition, \(y(0) = 0\), we can determine the value of \(C_2\): \[0 = \frac{1}{2}g(0)^2 + C_2\] \[C_2 = 0\] Thus, the displacement equation for the object is: \[y(t) = \frac{1}{2}gt^2\]

Step 5: Determine the time when the object hits the ground

To find the time when the object hits the ground, we will set the displacement equal to -100 meters (since the object is initially 100 meters above the ground): \[-100 = \frac{1}{2}gt^2\] Rearrange and solve for t: \[t^2 = \frac{-200}{g}\] In this problem, we assume the acceleration due to gravity \(g = 9.81 m/s^2\). Hence, \[t^2 = \frac{-200}{-9.81}\] \[t^2 = 20.39\] Now, we take the square root of both sides to get the time: \[t = \sqrt{20.39}\] \[t \approx 4.52 s\] So, the object will hit the ground after approximately 4.52 seconds.

Key Concepts

Differential Equations

A differential equation is a mathematical equation that relates some function with its derivatives. In practice, these functions usually represent physical quantities, and the derivatives represent their rates of change. The differential equation thus serves as a mathematical model that describes a physical process.
Differential equations can be classified by their order, determined by the highest derivative in the equation. For example, a first-order differential equation contains only the first derivative of the function, while a second-order differential equation contains, up to, the second derivative.
The simplest type of differential equations are linear, where the unknown function and its derivatives appear to the power of one and are not multiplied together. However, in the example provided, we can see a direct application: an object in free-fall is modeled by a second-order linear differential equation derived from Newton's second law of motion.

Second-order Differential Equation

A second-order differential equation deals with second derivatives and is vital in modeling many physical phenomena, such as oscillations and waves. For instance, the motion of a spring or simple harmonic motion can be described by second-order differential equations.
In our exercise, the equation \(\frac{d^2 y}{d t^2} = g\) is a second-order differential equation, which signifies that the acceleration (the second derivative of the position with respect to time) of the object is constant, equal to the acceleration due to gravity \(g\). This equation is also simple to solve because it is linear and homogeneous, meaning it can be solved by direct integration.

Integration of Differential Equations

The integration of differential equations is a fundamental method for finding solutions to differential equations. It involves finding the integral of both sides of the equation, which effectively 'reverses' the process of differentiation.
Integrating a first-order differential equation once will yield the original function, while integrating a second-order differential equation twice will do the same. With each integration step, a constant of integration is added, which is determined by the initial or boundary conditions given in the problem. As shown in the example problem, the constants were determined using the initial conditions that describe the object's velocity and position at time zero. The process of solving the equation by integrating it is part of what makes differential equations a powerful tool for understanding physical phenomena.

Free-fall Motion

The free-fall motion of an object is the motion of that object falling solely under the influence of gravity. This idealized form of motion does not take into account other forces like air resistance and is characterized by a uniform acceleration downward, equal to the standard gravitational acceleration on Earth, which is approximately \(9.81 m/s^2\).
The free-fall motion can be described using a second-order differential equation where the acceleration due to gravity is constant. By solving this second-order differential equation as we did in the exercise, we can model the object's displacement over time and find key information such as the time taken for the object to reach the ground, as seen in the final step of the provided solution.