Resources for Equation of Path
-
Questions
11
With Worked SolutionClick Here -
Video Tutorials
1
Click Here
Equation of Path Theory
![Changing from the parametric form of a projectiles trajectory \(v(t)=V t \cos \theta \,i+\left(V t \sin \theta-\frac{1}{2} g t^2\right) j\) to the cartesian form \(x=V t \cos \theta \quad y=V t \sin \theta-\frac{1}{2} g t^2.\)\\[4pt] Let \(t=\dfrac{x}{V \cos \theta}\) and then substitute into \(y=V t \sin \theta-\dfrac{1}{2} g t^2\). The equation is \( y=x \tan \theta-\dfrac{g}{2 V^2} x^2 \sec ^2 \theta \text {. } \)\\ \textbf{Example}\\ %30733 A body is projected so that on its upward path it passes through a point situated \(x_1\) metres horizontally and \(h\) metres vertically from the point of projection. The range is \(R\) metres and the angle of projection is \(\alpha\). The position of the particle is given by \(x = Vt\cos\alpha \qquad y = Vt\sin\alpha -\frac{1}{2}gt^2.\)\\ Where \(t\) is time, in seconds, after firing, and \(g\) is the acceleration due to gravity. \begin{center} \resizebox{0.5\textwidth}{!}{ \begin{tikzpicture} \def\angle{50} % starting angle \def\range{10} % horizontal range \def\height{0} % height \draw[-latex,line width=2pt] (0,0) -- (\range+2,0) node(X)[right] {\Large$x$}; \draw[-latex,line width=2pt] (0,0) -- (0,\height+4) node(Y)[above] {\Large$y$}; \draw[line width=2pt,->] (0,-1) -- (\range,-1) node[draw=gray!10!white,midway,rectangle,fill=gray!10!white] {\Large $R$}; \draw[line width=2pt,->] (0,-0.5) -- (\range-2,-0.5) node[draw=gray!10!white,midway,rectangle,fill=gray!10!white] {\Large$\displaystyle x_1$}; \def\v0{10} % initial velocity \def\g{9.8} % acceleration due to gravity \def\tflight{2*\v0*sin(\angle)/\g} % time of flight \def\xmax{\v0*\tflight*cos(\angle)} % horizontal range \draw[line width=2pt,dashed, samples=200, smooth,domain=0:\xmax] plot ({\x},{\height+\v0*sin(\angle)*(\x/\v0/cos(\angle))-0.5*\g*(\x/\v0/cos(\angle))^2}) node(G) {}; \draw[line width=2pt,dashed] (\range-2,0)--(\range-2,2) node[midway,left] {\Large $h$} node[above=5pt] {\Large $\alpha$}; \draw[fill=black] (\range-2,1.97) circle (0.12); \coordinate[label=above left:\Large$O$] (O) at (0,0); \coordinate (A) at (1,1); \pic [draw=black,line width=1pt,angle radius=0.9cm,angle eccentricity=1.5,"\Large$\alpha$"] {angle=X--O--A}; \end{tikzpicture} } %\includegraphics[width=0.35\textwidth]{e7a2bc50-1040-4649-9126-8623e92d9e08} \end{center} \begin{itemize}[nosep] \item[\bf{i)}]Show that the path of the trajectory is \(y = x\tan\alpha - \dfrac{gx^2}{2V^2\cos^2\alpha}.\) \item[\bf{ii)}]Show that \(V^2\cos^2\alpha = \dfrac{gR}{2\tan\alpha}.\) \item[\bf{iii)}]Show that \(\tan\alpha = \dfrac{Rh}{x_1(R-x_1)}.\) \end{itemize} \begin{multicols}{2} \textbf{Solution}\\ \textbf{i)} \; $\begin{aligned}[t]x &= Vt\cos\alpha \Rightarrow t = \frac{x}{V\cos\alpha}\\ \end{aligned}$\\ $\begin{aligned} y &= (V\sin\alpha)t - \frac{1}{2}gt^2\\&= (V\sin\alpha)\frac{x}{V\cos\alpha} - \frac{1}{2}g\frac{x^2}{V^2\cos^2\alpha}\\ \end{aligned}$\\ $\begin{aligned} \therefore y &= x\tan\alpha - \frac{gx^2}{2V^2\cos^2\alpha}.\end{aligned}$ \\ \\ \textbf{ii)}\; When \(y = 0\) and \(x = R\),\\ \(R\tan\alpha - \dfrac{gR^2}{2V^2\cos^2\alpha} = 0\)\\ $\begin{aligned}\frac{gR^2}{2V^2\cos^2\alpha} &= R\tan\alpha\\\frac{2V^2\cos^2\alpha}{gR^2} &= \frac{1}{R\tan\alpha}\\ \end{aligned}$\\ $\begin{aligned} V^2\cos^2\alpha &= \frac{gR^2}{2R\tan\alpha}\\\therefore V^2\cos^2\alpha &= \frac{gR}{2\tan\alpha}\end{aligned}$ \\ \\ \textbf{iii)} When \(x = x_1\) and \(y = h\),\\ $\begin{aligned}h &= x_1\tan\alpha - \frac{g{x_1}^2}{2V^2\cos^2\alpha}\\&= x_1\tan\alpha - g{x_1}^2\left(2\times\frac{gR}{2\tan\alpha}\right)^{-1}\\&= x_1\tan\alpha - \frac{{x_1}^2\tan\alpha}{R}\\ \end{aligned}$\\ $\begin{aligned} Rh &= x_1R\tan\alpha-{x_1}^2\tan\alpha\\Rh &= \tan\alpha(x_1R-{x_1}^2)\\\therefore \tan\alpha &= \frac{Rh}{x_1(R-x_1)}\end{aligned}$ \\ \end{multicols}](/media/yxwfg52x/equation-of-path.png)
NSW Y12 Maths - Extension 1 Projectile Motion Equation of Path
Videos
Videos relating to Equation of Path.
-
Equation of Path - Video - Projectiles - Cartesian equation of trajectory
You must be logged in to access this resource
Plans & Pricing
With all subscriptions, you will receive the below benefits and unlock all answers and fully worked solutions.
-
Teachers TutorsFeaturesFreeProAll ContentAll courses, all topicsQuestionsAnswersWorked SolutionsSystemYour own personal portalQuizbuilderClass ResultsStudent ResultsExam RevisionRevision by TopicPractise ExamsAnswersWorked SolutionsAwesome StudentsFeaturesFreeProContentAny course, any topicQuestionsAnswersWorked SolutionsSystemYour own personal portalBasic ResultsAnalyticsStudy RecommendationsExam RevisionRevision by TopicPractise ExamsAnswersWorked Solutions
Books / e-books
Course Book
Purchase the course book. This book either comes as a physical book or it can be purchased as an e-book.
Topic Books
You may choose to purchase the individual topic book from the main coursebook. These only come as e-books.
