Talk:PlanetPhysics/Projectile Motion

Original TeX Content from PlanetPhysics Archive

%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: projectile motion %%% Primary Category Code: 40. %%% Filename: ProjectileMotion.tex %%% Version: 3 %%% Owner: bloftin %%% Author(s): bloftin %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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\begin{document}

Consider the \htmladdnormallink{motion}{http://planetphysics.us/encyclopedia/CosmologicalConstant2.html} of a \htmladdnormallink{particle}{http://planetphysics.us/encyclopedia/Particle.html} which is projected in a direction making

an angle $\alpha$ with the \htmladdnormallink{horizon}{http://planetphysics.us/encyclopedia/GenericityInOpenSystems.html}. 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 \htmladdnormallink{Newton's laws of motion}{http://planetphysics.us/encyclopedia/NewtonsLaws.html} 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 y-axis} \\

$$ m \frac{d \ddot{y}}{dt} = -mg$$ \begin{equation} \frac{d \ddot{y}}{dt} = -g \end{equation}

where $\frac{d \ddot{x}}{dt}$ and $\frac{d \ddot{y}}{dt}$ are the components of the \htmladdnormallink{acceleration}{http://planetphysics.us/encyclopedia/Acceleration.html} 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 \htmladdnormallink{velocity}{http://planetphysics.us/encyclopedia/Velocity.html} 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 \htmladdnormallink{identity}{http://planetphysics.us/encyclopedia/Cod.html} 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 $\alpha = \frac{\pi}{4}$. 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 \htmladdnormallink{resistance}{http://planetphysics.us/encyclopedia/Conduction.html} 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.

\end{document}