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[parent] example constant acceleration and speed of sound (Example)

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?

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

We can either begin with the memorized constant acceleration equation

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

or start with the 2nd law of Newton's laws of motion

$\displaystyle F = ma $

Neglecting drag, the only force acting on the watermelon is gravity

$\displaystyle F = mg $

so

$\displaystyle ma = mg $

$\displaystyle 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 motion is

$\displaystyle \ddot{y} = g $

Integrating to get velocity

$\displaystyle \dot{y} = gt + C_1 $

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

$\displaystyle \dot{y} = gt $

Integrating to get position yields

$\displaystyle y = \frac{1}{2}gt^2 + C_2 $

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

$\displaystyle y = \frac{1}{2} g t^2$ (1)

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 $t_s$

$\displaystyle T = t + t_s $

Assuming the speed 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

$\displaystyle h = v_s t_s $

so

$\displaystyle t_s = \frac{h}{v_s} $

So $t$ in eq. 1 is then

$\displaystyle t = T - t_s = T - \frac{h}{v_s} $

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

$\displaystyle h = \frac{1}{2} g \left (T - \frac{h}{v_s} \right )^2$ (2)

Expanding

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

Collecting terms

$\displaystyle \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 quadratic formula to solve for $h$

$\displaystyle h = \frac{ -b \pm \sqrt{ b^2 - 4ac}}{2a} $

with the following constants

$\displaystyle a = \frac{g}{2 v_s^2} = \frac{9,8}{(2)(342)^2} = 4.1649313 \times 10^{-5} $

$\displaystyle b = \left ( -\frac{T}{v_s} - 1 \right ) = -\frac{2.9}{343} - 1 = 1.0084548 $

$\displaystyle 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,

$\displaystyle h_1 = 40.93 m $

and

$\displaystyle 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

$\displaystyle t = T - \frac{h}{v_s} $

For solution 1, we get

$\displaystyle t_1 = 2.78 s $

and for solution 2, we get

$\displaystyle 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

$\displaystyle h = 40.93 m $

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

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

which is only off by 0.7%.



"example constant acceleration and speed of sound" is owned by bloftin.

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Cross-references: quadratic formula, speed, position, velocity, motion, Newton's laws of motion, acceleration
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This is version 1 of example constant acceleration and speed of sound, born on 2007-08-12.
Object id is 262, canonical name is ExampleConstantAccelerationAndSpeedOfSound.
Accessed 2499 times total.

Classification:
Physics Classification: 45.50.Dd (General motion)
 45.50.Pk (Celestial mechanics )
 45.50.-j (Dynamics and kinematics of a particle and a system of particles)
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