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What is the altitude of the plane?

  1. Apr 10, 2008 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    Mach = vsource / vsound

    3. 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!
     
  2. jcsd
  3. Apr 10, 2008 #2
    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.
     
  4. Apr 10, 2008 #3
    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.
     
    Last edited: Apr 10, 2008
  5. Apr 10, 2008 #4

    tiny-tim

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    … one step at a time …

    Hi jrzygrl! :smile:

    No no no …

    All you've calculated is the horizontal distance travelled 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:
     
  6. Apr 10, 2008 #5

    Dick

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    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?
     
    Last edited: Apr 10, 2008
  7. Apr 10, 2008 #6
    Why would you do that? D=VT works just fine for this situation.
     
  8. Apr 10, 2008 #7

    tiny-tim

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    … one step at a time …

    Wrong! Do it step by step … :smile:
     
  9. Apr 10, 2008 #8

    Dick

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    It is so. :) I defined the triangle that way.
     
  10. Apr 10, 2008 #9

    Dick

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    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 travelling FASTER than sound. It's a shock wave.
     
  11. Apr 10, 2008 #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 im 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.
     
    Last edited: Apr 10, 2008
  12. Apr 10, 2008 #11

    Dick

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    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.
     
  13. Apr 10, 2008 #12
    thanks a lot guys i found the answer!
     
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