Talk:PlanetPhysics/Example Constant Acceleration and Speed of Sound

Original TeX Content from PlanetPhysics Archive
%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: example constant acceleration and speed of sound %%% Primary Category Code: 45.50.Dd %%% Filename: ExampleConstantAccelerationAndSpeedOfSound.tex %%% Version: 1 %%% 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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% 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}

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\begin{document}

A favorite past time on the David Letterman show is to throw watermelons off the roof of the Ed Sullivan theater. Let us determine the height of the building from the time $T$ of releasing the watermelon to hearing it splatter its contents on the street. Assuming $T = 2.9 s$. What is the height $h$ of the building?

\begin{center} \includegraphics[scale=.8]{ExConstantAccSpeedSound.eps} \end{center}

We can either begin with the memorized constant \htmladdnormallink{acceleration}{http://planetphysics.us/encyclopedia/Acceleration.html} equation

$$ x - x_0 = \frac{1}{2} \left( v_0 + v \right ) t $$

or start with the 2nd law of \htmladdnormallink{Newton's laws of motion}{http://planetphysics.us/encyclopedia/Newtons3rdLaw.html} $$ F = ma $$

Neglecting drag, the only \htmladdnormallink{force}{http://planetphysics.us/encyclopedia/Thrust.html} acting on the watermelon is gravity

$$ F = mg $$

so

$$ ma = mg $$ $$ a = g $$

Note that the acceleration is positvie because we chose the y-axis to be positive pointing down toward the street. So the equation of \htmladdnormallink{motion}{http://planetphysics.us/encyclopedia/CosmologicalConstant.html} is

\[ \ddot{y} = g   \]

Integrating to get \htmladdnormallink{velocity}{http://planetphysics.us/encyclopedia/Velocity.html} \[ \dot{y} = gt + C_1 \]

Since at $t = 0$, $\dot{y} = 0$, $C_1 = 0$ and therefore

\[ \dot{y} = gt \]

Integrating to get \htmladdnormallink{position}{http://planetphysics.us/encyclopedia/Position.html} yields

\[ y = \frac{1}{2}gt^2 + C_2 \]

Since at $t =0$, $y = 0$, we get

\begin{equation} y = \frac{1}{2} g t^2 \end{equation}

Although this part is straight forward, the trick for this problem is to incorporate the time needed for sound to travel from the street to our ears on the roof. Because $T$ is the total time, it includes the $t$ in eq. 1 and the time for sound to travel $t_s$

\[ T = t + t_s \]

Assuming the \htmladdnormallink{speed}{http://planetphysics.us/encyclopedia/Velocity.html} of sound $v_s$ at $20^o C$ is $343 m/s$, the height traveled in the time $t_s$ that must be subtracted off is

\[ h = v_s t_s \]

so

\[ t_s = \frac{h}{v_s} \]

So $t$ in eq. 1 is then

\[ t = T - t_s = T - \frac{h}{v_s} \]

Therefore, the height $(y = h)$ is now determined from

\begin{equation} h = \frac{1}{2} g \left (T - \frac{h}{v_s} \right )^2 \end{equation}

Expanding

\[ h = \frac{1}{2} g \left [ T^2 - \frac{2 T h}{v_s} + \frac{h^2}{v_s^2} \right ] \]

Collecting terms

\[ \frac{g}{2 v_s^2} (h^2) + \left ( -\frac{T}{v_s} - 1 \right ) (h) + \frac{1}{2} g T^2 = 0 \]

Now we can use the \htmladdnormallink{quadratic formula}{http://planetphysics.us/encyclopedia/QuadraticFormula.html} to solve for $h$

\[ h = \frac{ -b \pm \sqrt{ b^2 - 4ac}}{2a} \]

with the following constants

\[ a = \frac{g}{2 v_s^2} = \frac{9,8}{(2)(342)^2} = 4.1649313 \times 10^{-5} \] \[ b = \left ( -\frac{T}{v_s} - 1 \right )  = -\frac{2.9}{343} - 1 = 1.0084548 \] \[ c = \frac{1}{2} g T^2  = (0.5)(9.8)(2.9) = 41.209 \]

Substituting these values into the quadratic equation yields two solutions,

\[ h_1 = 40.93 m \]

and

\[ h_2 = 24172.07 m \]

To determine the correct answer we need to plug each solution into our time equation to see which one makes sense

\[ t = T - \frac{h}{v_s} \]

For solution 1, we get

\[ t_1 = 2.78 s \]

and for solution 2, we get

\[ t_2 = 67.57 s \]

Clearly, since our total time is $T = 2.9s$, only the first solution is correct and the height of the building is

\[ h = 40.93 m \]

Note, that if we ignored the speed of sound, we would use the equation

\[ h = \frac{1}{2} g t^2 = (0.5)(9.8)(2.9)^2 = 41.21 m \]

which is only off by 0.7\%.

\end{document}