Resources for Time Equations
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Video Tutorials
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Time Equations Theory
![A projectiles' trajectory can be defined as \\ \(\displaystyle v(t)=V t \cos \theta \,i+\left(V t \sin \theta-\frac{1}{2} g t^2\right)\, j\) in component form. \\[6pt] OR\\[6pt] \(v(t)=\left(\begin{array}{l} v t \cos \theta \\ v t \sin \theta-\dfrac{1}{2} g t^2 \end{array}\right) \text { in column vector form } \)\\ i.e. \(x=V t \cos \theta\) and \(y=V t \sin \theta-\dfrac{1}{2} g t^2\)\\[6pt] The velocity of the projectile can be defined as \(v(t)=v \cos \theta i+(v \sin \theta-g t) j\) in component form\\[6pt] OR\\[6pt] \(v(t)=\left(\begin{array}{l}v \cos \theta \\ v \sin \theta-g t\end{array}\right)\) in column vector form i.e \(\dot{x}=V \cos \theta\) and \(\dot{y}=V \sin \theta-g t\)\\[18pt] \textbf{Example}\\ A projectile is fired with initial velocity \(V\text{ms}^{-1}\) at an angle of projection \(\theta\) from a point \(O\) on horizontal ground. After 2 seconds it just passes over a 4 metre high wall that is 24 metres from the point of projection. Assume acceleration due to gravity is \(10\text{ms}^{-2}\). Assume the equations of displacement are $$x = Vt\cos\theta\text{ and } y = Vt\sin\theta - 5t^2$$ \begin{center} \resizebox{0.4\textwidth}{!}{ \begin{tikzpicture}[scale=0.85] \def\angle{45} % starting angle \def\range{12} % horizontal range \def\height{0} % height \draw[-latex,line width=2pt] (0,0) -- (\range,0) node(X)[right] {\Large$x$}; \draw[-latex,line width=2pt] (0,0) -- (0,\height+8) node(Y)[above] {\Large$y$}; \def\v0{12*sqrt(2)} % 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=3pt,dashed, samples=200, smooth,domain=0:%\xmax 11] plot ({\x},{\height+\v0*sin(\angle)*(\x/\v0/cos(\angle))-0.5*\g*(\x/\v0/cos(\angle))^2}) node(G) {}; \draw[line width=3pt] (\range-1,0)--(\range-1,12*1.2-5*1.2*1.2-1.6) node[midway,left=2pt] {\Large \textbf{4m}}; \path (0,0)--(\range-1,0) node[below=4pt,midway] {\Large \textbf{24m}}; \draw[fill=black] (\range-1,12*1.2-5*1.2*1.2-1.6) circle (0.2); \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} } \end{center} \begin{itemize} \item[\bf{i)}]Find \(V\) and \(\theta\). \item[\bf{ii)}]Find the time taken for the projectile to attain its maximum height. \item[\bf{iii)}]Find the range of the projectile. \end{itemize} \begin{multicols}{2} \textbf{Solution}\\ i)\\ When \(t=2\), \(x=24\),\\ $\begin{aligned}24 &= 2V\cos\theta \\V\cos\theta &= 12 \qquad (1)\end{aligned}$ \\ When \(t=2, y=4\)\\ $\begin{aligned}4 &= 2V\sin\theta -20\\V\sin\theta &= 12 \qquad (2)\end{aligned}$ \\ \((2) \div (1)\)\\ $\begin{aligned}\frac{V\sin\theta}{V\cos\theta} &= \tan\theta = 1\\\therefore \theta &= \frac{\pi}{4}\end{aligned}$ \\ $\begin{aligned}x &= Vt\cos\theta\\24 &= V\times\frac{1}{\sqrt{2}}\times 2\\\therefore V &= 12\sqrt{2}\text{ms}^{-1}\end{aligned}$ \\ ii)\\ $\begin{aligned}y &= Vt\sin\theta - 5t^2\\\dot{y} &= V\sin\theta -10t\end{aligned}$ \\ Let \(\dot{y}\) equal \(0\) to determine the time for the particle to attain maximum height,\\ $\begin{aligned}0 &= 12\sqrt{2}\times\frac{1}{\sqrt{2}} - 10t\\10t &= 12\\\therefore t &= 1.2 \text{ seconds}\end{aligned}$ \\ iii)\\ If the time to reach the maximum height is 1.2 seconds, then by symmetry the time taken for the particle to strike the ground is 2.4 seconds.\\ $\begin{aligned}x &= Vt\cos\theta\\&= 12\sqrt{2}\times\frac{1}{\sqrt{2}}\times 2.4= 28.8\text{m}\end{aligned}$ \\ \end{multicols}](/media/2gkjjvpt/4853.png)
NSW Y12 Maths - Extension 1 Projectile Motion Time Equations
Videos
Videos relating to Time Equations.
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Time Equations - Video - Projectile Practice Problem 1
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Time Equations - Video - Projectile Practice Problem 2
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Time Equations - Video - Projectile Practice Problem 3
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Time Equations - Video - Projectile Practice Problem 4
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Time Equations - Video - Projectile Practice Problem 5
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