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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
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Equation of Path - Video - Projectiles - Cartesian equation of trajectory
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