What is the altitude of the plane?

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Homework Help Overview

The problem involves a jet fighter flying at a speed corresponding to Mach 1.29, with the sonic boom heard by an observer on the ground 11.9 seconds later. The task is to determine the altitude of the plane, assuming the speed of sound remains constant regardless of altitude.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the validity of the speed of sound used in calculations, questioning its dependence on environmental factors. There are debates about the correct application of distance formulas and the interpretation of the sonic boom's propagation. Some suggest drawing a right triangle to visualize the relationship between the plane's altitude and the distance traveled by sound.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Participants are actively questioning assumptions and clarifying the setup of the problem. Some guidance has been offered regarding the geometric representation of the scenario, but no consensus has been reached.

Contextual Notes

There are indications that participants are grappling with the implications of the sonic boom's behavior and the assumptions about sound travel in relation to the plane's altitude. The original poster's calculations have been challenged, leading to further exploration of the problem's parameters.

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Homework Statement



A jet fighter in level flight passes directly overhead at a speed corresponding to Mach number 1.29. The sonic boom is heard by you on the ground 11.9 s later. What is the altitude of the plane? Assume (unrealistically) the speed of sound does not change with altitude.

Homework Equations



Mach = vsource / vsound

The Attempt at a Solution



M = vsource / vsound
1.29 = vsource / 343
vsource = 442.47 m/s

d = vt
d = (442.47 m/s)(11.9 s)
d = 5265.393m = 5.265 km

the answer above was wrong, any help is appreciated!
 
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Ok, well, to start, are you sure that 343m/s is the correct number? Because it varies depending on temperature, humidity and many other factors. It would probably be helpful if it was one of the givens.
 
And why would you put the speed of the jet into D=VT? You put T as the time it took the SOUND to reach you from the point it was exerted(which we will call point A), so you should probably put V as the speed of sound. This will tell you how far away Point A is, which will tell you how high up the plane was.
 
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… one step at a time …

jrzygrl said:
d = vt
d = (442.47 m/s)(11.9 s)
d = 5265.393m = 5.265 km

Hi jrzygrl! :smile:

No no no …

All you've calculated is the horizontal distance traveled by sound in 11.9s.

How does that give you its height?

Think … do it one step at a time … can you hear the plane as it approaches? can you hear it when it's overhead (position O, say)? Where is it when you can first hear it (position F, say)? How long does the plane take to get from O to F? How long does the sound take to get from F to J (that's you)?

Then draw the triangle OFJ … :smile:
 
Draw a right triangle with you at one vertex, the plane at 11.9sec later at another and the place where the plane was when it was directly overhead at the right angle. If you just heard the boom then the hypotenuse is the shock wave. The horizontal leg is the velocity of the plane times 11.9sec, right? You want to solve for the vertical leg. To solve it by trig you need to know one more thing about the triangle. Can you use the given information to find the angle between the shock wave and the horizontal?
 
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Why would you do that? D=VT works just fine for this situation.
 
… one step at a time …

Dick said:
The horizontal leg is the velocity of the plane time 11.9sec, right?

Wrong! Do it step by step … :smile:
 
tiny-tim said:
Wrong! Do it step by step … :smile:

It is so. :) I defined the triangle that way.
 
KalvinDeathX said:
Why would you do that? D=VT works just fine for this situation.

If you are saying that the distance from you to the plane when you hear the boom is V_sound*11.9sec, you are quite wrong. The plane is traveling FASTER than sound. It's a shock wave.
 
  • #10
No, I'm saying the distance from you to the point directly above you at the height where the plane passed. I'm not saying I'm right though.
If a sonicboom works this way, then I am probably wrong -> upload.wikimedia.org/wikipedia/commons/b/b3/Sonic_boom.svg
I just assumed the sound might travel straight down, but now I think that it comes out in a cone, which would make trig necessary.
 
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  • #11
KalvinDeathX said:
No, I'm saying the distance from you to the point directly above you at the height where the plane passed.

Also untrue. You hear the sound when the shock front passes. The point on the shock front that hits you actually originated from the plane before it passed overhead. There's a picture at http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/sound/u11l3b.html .
 
Last edited by a moderator:
  • #12
thanks a lot guys i found the answer!
 

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