Problem 29 - Thomas Calculus

Question

Firing from \(\left(x_{0}, y_{0}\right)\) Derive the equations $$\begin{array}{l}{x=x_{0}+\left(v_{0} \cos \alpha\right) t} \\\ {y=y_{0}+\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2}}\end{array}$$ (see Equation \((7)\) in the text) by solving the following initial value problem for a vector \(r\) in the plane. $$\text{Differential equation:}\quad \frac{d^{2} \mathbf{r}}{d t^{2}}=-g \mathbf{j}$$ $$\text{Initial conditions:}\begin{array}{l}{\mathbf{r}(0)=x_{0} \mathbf{i}+y_{0} \mathbf{j}} \\ {\frac{d \mathbf{r}}{d t}(0)=\left(v_{0} \cos \alpha\right) \mathbf{i}+\left(v_{0} \sin \alpha\right) \mathbf{j}}\end{array}$$

Step-by-Step Solution

Answer

Use the differential equation and initial conditions to integrate for velocity and position.

Step 1: Identify the Physical Context

We are dealing with projectile motion in a two-dimensional plane, where the projectile is subject to gravitational acceleration only. The position vector \(\mathbf{r}\) is defined as \(\mathbf{r} = x \mathbf{i} + y \mathbf{j}\), where \(x\) and \(y\) denote the coordinates, and \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors in the horizontal and vertical directions, respectively.

Step 2: Apply the Differential Equation

The differential equation given by \( \frac{d^{2} \mathbf{r}}{dt^{2}} = -g \mathbf{j} \) indicates that the only acceleration is due to gravity. This means the horizontal acceleration is zero and the vertical acceleration is \(-g\).

Step 3: Integrate to Find Velocity

To find the velocity, integrate the acceleration equation \( \frac{d^{2} \mathbf{r}}{d t^{2}} = -g \mathbf{j} \). This gives us: \(\frac{d \mathbf{r}}{dt} = \mathbf{v} = C \mathbf{i} - gt \mathbf{j}\), where \(C\) is an integration constant for the \(\mathbf{i}\) component. Using the initial condition \( \frac{d \mathbf{r}}{d t}(0) = (v_0 \cos \alpha) \mathbf{i} + (v_0 \sin \alpha) \mathbf{j} \), we find \(v = (v_0 \cos \alpha) \mathbf{i} + (v_0 \sin \alpha - gt) \mathbf{j}\).

Step 4: Integrate to Find Position

Integrate the velocity equation \(\mathbf{v} = (v_0 \cos \alpha) \mathbf{i} + (v_0 \sin \alpha - gt) \mathbf{j}\) to find position \(\mathbf{r}\). This gives: \(\mathbf{r}(t) = \left( x_0 + (v_0 \cos \alpha) t \right) \mathbf{i} + \left( y_0 + (v_0 \sin \alpha) t - \frac{1}{2} gt^2 \right) \mathbf{j} \).

Step 5: Derive Final Position Equations

Extract the scalar equations from \(\mathbf{r}(t)\): \[x = x_0 + (v_0 \cos \alpha) t\] for the x-coordinate and \[y = y_0 + (v_0 \sin \alpha) t - \frac{1}{2} gt^2\] for the y-coordinate.

Key Concepts

Differential Equations

Differential equations play a crucial role in understanding projectile motion. They are equations that relate a function with its derivatives. In the context of projectile motion, the second-order differential equation is given by \( \frac{d^{2} \mathbf{r}}{dt^{2}} = -g \mathbf{j} \). This equation indicates the relationship between time and acceleration, specifically concerning gravity's effect.
The second derivative of the position vector \( \mathbf{r} \) with respect to time tells us about the projectile's acceleration. Since the only force acting on a projectile in motion (neglecting air resistance) is gravitational force, the only component of acceleration present is in the vertical direction, \(-g\). This fundamental setup simplifies the understanding of projectile motion, allowing us to derive the velocity and position of a projectile over time through further integrations.

Initial Value Problem

An initial value problem is a type of differential equation that comes with specific conditions. These initial conditions are essential because they specify the state of the system at a starting point, which in our case is time \( t=0 \).
For our problem, the initial conditions are \( \mathbf{r}(0) = x_{0} \mathbf{i} + y_{0} \mathbf{j} \) for position and \( \frac{d \mathbf{r}}{d t}(0) = (v_0 \cos \alpha) \mathbf{i} + (v_0 \sin \alpha) \mathbf{j} \) for velocity.

• These conditions allow us to solve the differential equations uniquely because they define where and how fast the projectile starts at \( t=0 \).
• They act as the baseline from which we determine both the velocity and the position of the projectile at any future time \( t \).

Solving these conditions involves integrating the differential equation to find expressions for velocity and subsequently for position.

Gravitational Acceleration

Gravitational acceleration, represented as \( g \), is a constant that describes the rate at which an object accelerates as it is pulled by the force of gravity toward the Earth. In projectile motion, this is often simplified to \( g = 9.8 \, m/s^2 \).
This acceleration acts downwards and only affects the vertical motion of the projectile, as seen in the equation \( \frac{d^{2} \mathbf{r}}{dt^{2}} = -g \mathbf{j} \).

• The minus sign indicates the downward direction of the gravitational force.
• Horizontal motion remains unaffected by gravity, leading to a constant horizontal velocity.

These distinctions help derive clear equations of motion for when a projectile is subjected to gravitational acceleration alone.

Vector Calculus

Vector calculus is an essential mathematical tool used to solve the projectile motion problem. It allows us to handle quantities like position, velocity, and acceleration as multi-dimensional vectors.
In our exercise, the position vector \( \mathbf{r} \) is decomposed into components \( x \mathbf{i} \) and \( y \mathbf{j} \), representing horizontal and vertical motions, respectively.

• Integrating the differential equation for acceleration, we use vector calculus to find the velocity vector \( \mathbf{v} \).
• Further integration yields the position vector \( \mathbf{r} \), consisting of an equation for each of the horizontal and vertical components.

Understanding vector calculus concepts like integration and vector decomposition is vital in deriving and working with the equations governing projectile motion.