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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 = 2.9s. What is the height h of the building?

PIC

We can either begin with the memorized constant acceleration equation

x x0 = 1 2 (v0 + v)t

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

F = ma

Neglecting drag, the only force 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 motion is

ÿ = g

Integrating to get velocity

= gt + C1

Since at t = 0, = 0, C1 = 0 and therefore

= gt

Integrating to get position yields

y = 1 2gt2 + C 2

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

y = 1 2gt2 (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 ts

T = t + ts

Assuming the speed of sound vs at 20oC is 343ms, the height traveled in the time ts that must be subtracted off is

h = vsts

so

ts = h vs

So t in eq. 1 is then

t = T ts = T h vs

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

h = 1 2g (T h vs ) 2 (2)

Expanding

h = 1 2g [T2 2Th vs + h2 vs2 ]

Collecting terms

g 2vs2(h2) + (T vs 1) (h) + 1 2gT2 = 0

Now we can use the quadratic formula to solve for h

h = b ±b2 4ac 2a

with the following constants

a = g 2vs2 = 9, 8 (2)(342)2 = 4.1649313 × 105
b = (T vs 1) = 2.9 343 1 = 1.0084548
c = 1 2gT2 = (0.5)(9.8)(2.9) = 41.209

Substituting these values into the quadratic equation yields two solutions,

h1 = 40.93m

and

h2 = 24172.07m

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

t = T h vs

For solution 1, we get

t1 = 2.78s

and for solution 2, we get

t2 = 67.57s

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

h = 40.93m

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

h = 1 2gt2 = (0.5)(9.8)(2.9)2 = 41.21m

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 2642 times total.

Classification:
Physics Classification45.50.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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