PlanetPhysics/Example Constant Acceleration and Speed of Sound

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/|acceleration]] equation

$x - x_{0} = \frac{1}{2} (v_{0} + v) t$

or start with the 2nd law of [[../Newtons3rdLaw/|Newton's laws of motion]] $F = m a$

Neglecting drag, the only [[../Thrust/|force]] acting on the watermelon is gravity

$F = m g$

so

$m a = m g$ $a = g$

Note that the acceleration is positvie because we chose the y-axis to be positive pointing down toward the street. So the equation of [[../CosmologicalConstant/|motion]] is

$\ddot{y} = g$

Integrating to get [[../Velocity/|velocity]]

$\dot{y} = g t + C_{1}$

Since at $t = 0$ , $\dot{y} = 0$ , $C_{1} = 0$ and therefore

$\dot{y} = g t$

Integrating to get [[../Position/|position]] yields

$y = \frac{1}{2} g t^{2} + C_{2}$

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

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

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 [[../Velocity/|speed]] of sound $v_{s}$ at $2 0^{o} C$ 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

$h = \frac{1}{2} g {(T - \frac{h}{v_{s}})}^{2}$

Expanding

$h = \frac{1}{2} g [T^{2} - \frac{2 T h}{v_{s}} + \frac{h^{2}}{v_{s}^{2}}]$

Collecting terms

$\frac{g}{2 v_{s}^{2}} (h^{2}) + (- \frac{T}{v_{s}} - 1) (h) + \frac{1}{2} g T^{2} = 0$

Now we can use the [[../QuadraticFormula/|quadratic formula]] to solve for $h$

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

with the following constants

$a = \frac{g}{2 v_{s}^{2}} = \frac{9, 8}{(2) (342)^{2}} = 4.1649313 \times 1 0^{- 5}$

$b = (- \frac{T}{v_{s}} - 1) = - \frac{2.9}{343} - 1 = 1.0084548$

$c = \frac{1}{2} g T^{2} = (0.5) (9.8) (2.9) = 41.209$

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 $T = 2.9 s$ , 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\

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PlanetPhysics/Example Constant Acceleration and Speed of Sound

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