ProjectileMotion 2006-08-06 20:19:24 2006-08-06 20:19:24 Definition Ballistics (2D) % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) \usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Consider the motion of a particle which is projected in a direction making an angle $\alpha$ with the horizon. When we neglect drag, the only force which acts upon the particle is its weight, $m{\bf g}$ (Fig. 66). \begin{figure} \includegraphics[scale=.8]{Fig66.eps} \end{figure} Taking the plane of motion to be the xy-plane, and applying Newton's laws of motion gives us the equations {\bf x-axis} \\ $$ m \frac{d \ddot{x}}{dt} = 0$$ \begin{equation} \frac{d \ddot{x}}{dt} = 0 \end{equation} {\bf x-axis} \\ $$ m \frac{d \ddot{y}}{dt} = -mg$$ $$ \frac{d \ddot{y}}{dt} = -g $$ where $\frac{d \ddot{x}}{dt}$ and $\frac{d \ddot{y}}{dt}$ are the components of the acceleration along the x and y axes. Integrating equations (1) and (2) we get $$ \dot{x} = c_1 $$ $$ \dot{y} = -gt + c_2 $$ Therefore the component of the velocity along the x-axis remains constant, while the component along the y-axis changes uniformly. Let $v_0$ be the initial velocity of the projection, then when $t = 0$, $\dot{x}_0 = v_0 \cos \alpha$ and $\dot{y}_0 = v_0 \sin \alpha$. Making these substitutions in the last two equations we obtain $$ c_1 = v_0 \cos \alpha $$ $$ c_2 = v_0 \sin \alpha $$ Therefore \begin{equation} \dot{x} = v_0 \cos \alpha \end{equation} \begin{equation} \dot{y} = v_0 \sin \alpha - gt \end{equation} Then the total velocity at any instant is $$ v = \sqrt{\dot{x}^2 + \dot{y}^2} $$ $$ v = \sqrt{v_0^2 - 2v_0 g t \sin \alpha + g^2 t^2} $$ and makes an angle $\theta$ with the horizon defined by $$ \tan \theta = \frac{ \dot{y} }{ \dot{x} } = \frac{ v_0 \cos \alpha}{ v_0 \sin \alpha - gt} $$ Integrating equations (3) and (4) we obtain $$ x = v_0 t \cos \alpha + c_3 $$ $$ y = v_0 t \sin \alpha - \frac{1}{2} g t^2 + c_4 $$ But when $t = 0$, $x = y = 0 $, therefore $c_3 = c_4 = 0$, and consequently \begin{equation} x = v_0 t \cos \alpha \end{equation} \begin{equation} y = v_0 t \sin \alpha - \frac{1}{2} g t^2 \end{equation} It is interesting to note that the motions in the two directions are independent. The gravitational acceleration does not affect the constant velocity along the x-axis, while the motion along the y-axis is the same as if the body were dropped vertcally with an initial velocity $v_0 \sin \alpha $. {bf The Path} - The equation of the path may be obtained by eliminating $t$ between equations (7) and (8). This gives \begin{equation} y = x \tan \alpha - \frac{g}{2 v_0^2 \cos^2 \alpha} x^2 \end{equation} which is the equation of a parabola. {bf The Time of Flight} - When the projectile strikes the ground its y-coordinate is zero. Therefore substituting zero for $y$ in equation (8) we get for the time of flight \begin{equation} T = \frac{2 v_0 \sin \alpha}{g} \end{equation} {\bf The Range} - The range, or the total horizontal distance covered by the projectile, is found by replacing $t$ in equation (7) by the value of $T$ in equation (10), or by letting $y = 0$ in equation (9). By either method we obtain \begin{equation} R = \frac{2 v_0^2 \sin \alpha \cos \alpha}{g} = \frac{v_0^2 \sin 2 \alpha}{g} \end{equation} Note that a basic trigonometric identity was used to simpilfy the above equation. Since $v_0$ and $g$ are constants the value of $R$ depends upon $\alpha$. It is evident from equation (11) that $R$ is maximum when $\sin 2 \alpha = 1$, or when $\alpha = \frac{\pi}{2}$. The maximum range is, therefore, \begin{equation} R_{max} = \frac{v_0^2}{g} \end{equation} In actual practice the angle of elevation which gives the maximum range is smaller on account of the resistance of the air. {\bf The Highest Point} - At the highest point $\dot{y}=0$. Therefore substituting this value of $\dot{y}$ in equation (4) we obtain $\frac{v_0 \sin \alpha}{g}$ or $\frac{1}{2}T$ for the time taken to reach the highest point. Subsituting this value of the time in equation (8) we get for the maximum elevation \begin{equation} H = \frac{v_0^2 \sin^2 \alpha}{2 g} \end{equation} {\bf The Range for a Sloping Ground} - Let $\beta$ be the angle which the ground makes with the horizon. Then the range is the distance $OP$, Fig. 67, where $P$ is the point where the projectile strikes the sloping ground. The equation of the line $OP$ is \begin {equation} y = x \tan \beta \end{equation} \begin{figure} \includegraphics[scale=.8]{Fig67.eps} \end{figure} Eliminating $y$ between equations (14) and (9) we obtain the x-coordinate of the point, $$ x_p = \frac{ 2 v_0^2 \cos^2 \alpha \left ( \tan \alpha - \tan \beta \right ) }{g} $$ But $x_p = R' \cos \beta$, where $R' = OP$. Therefore $$ R' = \frac{2 v_0^2 \cos \alpha}{g \cos^2 \beta} \sin \left ( \alpha - \beta \right ) $$ \begin{equation} R' = \frac{ v_0^2}{g} \frac{ \sin \left ( 2 \alpha - \beta \right ) - \sin \beta }{ \cos^2 \beta } \end{equation} Thus for a given value of $\beta$, $R'$ is maximum when $\sin \left ( 2 \alpha - \beta \right ) = 1$, that is, when $\alpha = \frac{\pi}{4} + \frac{\beta}{2}$. \begin{equation} R_{max}' = \frac{ v_0^2 1 - \sin \beta}{ g \cos^2 \beta} = \frac{v_0^2}{g \left ( 1 + \sin \beta \right ) } \end{equation} When $\beta = 0$ equations (15) and (16) reduce to equations (12) and (13), as they should.