/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Firing from \(\left(x_{0}, y_{0}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Firing from \(\left(x_{0}, y_{0}\right)\) Derive the equations $$ \begin{aligned} x &=x_{0}+\left(v_{0} \cos \alpha\right) t \\ y &=y_{0}+\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2} \end{aligned} $$ (see Equation ( 5\()\) in the text) by solving the following initial value problem for a vector \(\mathbf{r}\) in the plane. $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d^{2} \mathbf{r}}{d t^{2}}=-g \mathbf{j}} \\ {\text { Initial conditions: }} & {\mathbf{r}(0)=x_{0} \mathbf{i}+y_{0} \mathbf{j}} \\ {} & {\frac{d \mathbf{r}}{d t}(0)=\left(v_{0} \cos \alpha\right) \mathbf{i}+\left(v_{0} \sin \alpha\right) \mathbf{j}}\end{array} $$

Short Answer

Expert verified
The equations of motion are derived as: \[ x(t) = x_0 + (v_0 \cos \alpha)t \] \[ y(t) = y_0 + (v_0 \sin \alpha)t - \frac{1}{2}gt^2 \]

Step by step solution

01

Interpret the Problem

We need to derive the equations of motion for a projectile given initial conditions. The task requires solving a second-order differential equation in two dimensions to find the position vector \(\mathbf{r}\). We'll begin with the acceleration due to gravity acting only in the \(y\)-direction, represented by \(-g \mathbf{j}\).
02

Separate the Differential Equation

The differential equation given is \(\frac{d^{2} \mathbf{r}}{d t^{2}} = -g \mathbf{j}\). This means that the acceleration in the \(x\)-direction is zero, and in the \(y\)-direction, it is \(-g\). We need to solve this for both \(x\) and \(y\) as components of \(\mathbf{r}\).
03

Solve for x-component

Since \(\frac{d^{2} x}{d t^{2}} = 0\), integrate to find \(\frac{d x}{d t} = C_1\). Using the initial condition \(\frac{d x}{d t}(0) = v_0 \cos \alpha\), we find \(C_1 = v_0 \cos \alpha\). Integrate again to get \(x(t) = v_0 \cos \alpha \cdot t + C_2\). Applying the initial position \(x(0) = x_0\), it follows that \(C_2 = x_0\). Thus, \(x(t) = x_0 + (v_0 \cos \alpha)t\).
04

Solve for y-component

For the \(y\)-component, solve \(\frac{d^2 y}{d t^2} = -g\). Integrate once to get \(\frac{d y}{d t} = -gt + C_3\). Using the initial condition \(\frac{d y}{d t}(0) = v_0 \sin \alpha\), we find \(C_3 = v_0 \sin \alpha\). Integrate again to get \(y(t) = -\frac{1}{2}gt^2 + v_0 \sin \alpha \cdot t + C_4\). Applying the initial position \(y(0) = y_0\), it follows that \(C_4 = y_0\). Thus, \(y(t) = y_0 + (v_0 \sin \alpha)t - \frac{1}{2}gt^2\).
05

Combine Derived Equations

Combine \(x(t) = x_0 + (v_0 \cos \alpha)t\) and \(y(t) = y_0 + (v_0 \sin \alpha)t - \frac{1}{2}gt^2\) to represent the position vector \(\mathbf{r}(t)\). These equations, derived from the initial value problem, describe the projectile's motion.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Differential Equations in Projectile Motion
A differential equation is a mathematical expression that relates a function with its derivatives. In projectile motion, differential equations help us describe how various physical quantities change over time.
For projectiles, like a tossed ball, the motion according to Newton's laws is summarized with differential equations. They consist of initial velocity, gravity, and resistance (if any). In this context, understand that gravity affects the projectile only in the vertical direction (denoted by the vector \(-g \mathbf{j}\)). It indicates that the horizontal motion (x-direction) is uniform, meaning there's no acceleration from external forces in that direction.
By solving differential equations, specifically second-order for motion problems, we can derive expressions for the position and velocity of the projectile as functions of time. In balance with initial conditions, the differential equations yield specific formulas for the trajectory of the projectile.
Understanding the Initial Value Problem
When solving dynamics problems like projectile motion, we often encounter initial value problems (IVP). This is a specific type of differential equation problem where initial conditions are given. For example, a projectile's starting position and initial velocity vector.
In this problem, you start with conditions like the initial location, \(\mathbf{r}(0) = x_0 \mathbf{i} + y_0 \mathbf{j}\), and initial velocities along each axis. These stipulations allow the solving of differential equations to obtain precise motion equations. IVPs are crucial as they define the starting point for the motion and thereby influence subsequent calculations.
Solving an IVP enables us to track how the projectile moves over time. With known initial speeds and directions, combined with the gravitational pull, we can predict the complete trajectory, determining both how far and how high the projectile will go.
Vector Calculus for Analyzing Motion
Vector calculus involves using vectors to study physical quantities that have magnitude and direction, a key area in projectile motion analysis. In this context, displacement, velocity, and acceleration are considered as vectors, enabling better representation and computation.
The position vector \(\mathbf{r}\) represents the projectile's location in a two-dimensional space, \(x\text{-direction}\) and \(y\text{-direction}\). Breaking down these components helps us focus separately on how the projectile travels horizontally and vertically. When analyzing motion, recognizing that vectors simplify these complex movements by providing clarity to direction makes calculation and understanding more intuitive.
Calculating the derivatives of vectors, one finds velocities and accelerations, laying out how positions change over time. This method proves especially helpful in projectile motion where different forces may act in varied directions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 16 and \(17,\) two planets, planet \(A\) and planet \(B\) , are orbiting their sun in circular orbits with \(A\) being the inner planet and \(B\) being farther away from the sun. Suppose the positions of \(A\) and \(B\) at time \(t\) are $$ \mathbf{r}_{A}(t)=2 \cos (2 \pi t) \mathbf{i}+2 \sin (2 \pi t) \mathbf{j} $$ and $$ \mathbf{r}_{B}(t)=3 \cos (\pi t) \mathbf{i}+3 \sin (\pi t) \mathbf{j} $$ respectively, where the sun is assumed to be located at the origin and distance is measured in astronomical units. (Notice that planet \(A\) moves faster than planet \(B . )\) The people on planet \(A\) regard their planet, not the sun, as the center of their planetary system (their solar system). Using planet \(A\) as the origin, graph the path of planet \(B .\) This exercise illustrates the difficulty that people before Kepler's time, with an earth- centered (planet \(A\) ) view of our solar system, had in understanding the motions of the planets (i.e., planet \(B=\) Mars). See D. G. Saari's article in the American Mathematical Monthly, Vol. 97 (Feb. \(1990 ),\) pp. \(105-119\) .

Hitting a baseball under a wind gust A baseball is hit when it is 2.5 \(\mathrm{ft}\) above the ground. It leaves the bat with an initial velocity of 145 \(\mathrm{ft} / \mathrm{sec}\) at a launch angle of \(23^{\circ} .\) At the instant the ball is hit, an instantaneous gust of wind blows against the ball, adding a component of \(-14 \mathbf{i}(\mathrm{ft} / \mathrm{sec})\) to the ball's initial velocity. A 15 -ft-high fence lies 300 \(\mathrm{ft}\) from home plate in the direction of the flight. a. Find a vector equation for the path of the baseball. b. How high does the baseball go, and when does it reach maximum height? c. Find the range and flight time of the baseball, assuming that the ball is not caught. d. When is the baseball 20 \(\mathrm{ft}\) high? How far (ground distance) is the baseball from home plate at that height? e. Has the batter hit a home run? Explain.

In Exercises \(1-8,\) find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$ \mathbf{r}(t)=(t \cos t) \mathbf{i}+(t \sin t) \mathbf{j}+(2 \sqrt{2} / 3) t^{3 / 2} \mathbf{k}, \quad 0 \leq t \leq \pi $$

In Exercises \(1-8,\) find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$ \mathbf{r}(t)=(t \sin t+\cos t) \mathbf{i}+(t \cos t-\sin t) \mathbf{j}, \quad \sqrt{2} \leq t \leq 2 $$

In Exercises 9 and \(10,\) write a in the form \(a_{\mathrm{T}} \mathbf{T}+a_{\mathrm{N}} \mathbf{N}\) without finding \(\mathbf{T}\) and \(\mathbf{N} .\) $$ \mathbf{r}(t)=(1+3 t) \mathbf{i}+(t-2) \mathbf{j}-3 t \mathbf{k} $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.