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Homework Help: Sonic boom plane altitude problem

  1. Apr 14, 2006 #1
    Hi, I am having trouble with the following question:

    A jet fighter in level flight passes directly overhead at a speed corresponding to Mach number 1.35. The sonic boom is heard by you on the ground 12.0 s later. What is the altitude of the plane? Assume (unrealistically) the speed of sound does not change with altitude. Answer should be in km.

    I know that Mach numbers are velocity(object)/velocity(medium) which gives:
    1.35 = velocity(object)/343 m/s, therefore velocity of the plane is 463.05 m/s

    From there I took the time, 12.0s, and multiplied by 463.05m/s to get 5556.6 m, or 5.5566km, but this is not right.

    COuld anyone offer me some suggestions for this one? Thanks!!!!!
  2. jcsd
  3. Apr 14, 2006 #2


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    Why can't you simply use;

    [tex]v = \frac{ds}{dt}[/tex]

  4. Apr 14, 2006 #3
    sorry, I am not familiar with how to use that kind of equation, could you explain what it means, please? thanks!
  5. Apr 14, 2006 #4


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    I meant, why does the velocity of the aircraft matter in how long it takes the sound to reach the ground?

  6. Apr 14, 2006 #5
    do you mean that i should just use the velocity of the speed of sound?
    so, 343 m/s * 12.0 s = 4116 m = 4.116 km ?

    i was thinking the velocity of the plane does matter b/c since it is above mach 1, it makes a cone shaped wave, but does this not matter?

    thank you!
  7. Apr 14, 2006 #6


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    It is this cone (or the rapid pressure change between the base of the cone and the atmospheric pressure) that creates the sonic boom. Your answer appears correct.

  8. Apr 14, 2006 #7
    ok thank you, although apparently 4.116 km is not right either so i am still confused about this one...
  9. Apr 14, 2006 #8
    I believe only the velocity of light has the unique property of being constant no matter at what velocity the observer or the listener are moving.
    Correct me if I am wrong, but I feel velocity of sound is affected by the speed of the aircraft.

    - Arun
  10. Apr 14, 2006 #9
    so any ideas on how to factor in the speed of the aircraft with the speed at which the sound reaches the observer?
  11. Apr 14, 2006 #10

    Doc Al

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    Hints & Suggestions:

    (1) What angle does the sonic cone make with the horizontal?

    (2) The moment you hear the sound is the moment the cone hits you.

    (3) Consider the triangle formed by you, the point overhead, and the point where the plane is at the moment you hear the boom.
    Last edited: Apr 14, 2006
  12. Apr 14, 2006 #11


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    Ahh, it all makes sense now!!That's why you need the velcoity of the aircraft!! That's been doing my head in for the past hour! :grumpy: Thanks for answering Doc Al you've put my mind at rest :approve:

  13. Apr 14, 2006 #12
    so the angle of the cone is found by: sin(theta)=1/Mach#=.7407, so theta = 47.8degrees (approx). since the plane is flying horizontal, divide this angle by 2 and get 23.9degrees (1)

    (3)triangle is a right triangel with sides as distances from "you" to point overhead and point overhead to point where plane is when "you" hear the sound (that's the right angle). the angle from "you" to the plane when you hear it is 90-23.9=66.1degrees.
    using the equation tan(theta)=opp./adjacent
    i can fill in tan(66.1)=55566m/x and solve for x and get:
    x=24620m or 24.6km

    am i understanding this correctly? if so, does my math look correct, b/c i am still getting the wrong answer...

  14. Apr 14, 2006 #13


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    Phoey. I don't get it. My little Post-It diagram still has the vertical component of the velocity diagram at mach 1. The horizontal is the plane at mach 1.35, but I still get mach 1 for 12 seconds for the vertical displacement. Any more hints for a Friday-impaired brain?
  15. Apr 14, 2006 #14
    anyone know if i am picturing the triangle correctly?
  16. Apr 14, 2006 #15

    Doc Al

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    Why did you divide by 2?
  17. Apr 14, 2006 #16
    should i not divide by two?
    it is difficult to explain why i thought i should in writing but i'll try:

    so the cone is horizontal to the ground (on its side, for a lack of a better way of explaining it), with the imaginary parallel horizon going through the cone, splitting it in half, sort of like if you split an equilateral triangle down the middle from one point to the opposite side how you get two right triangles. so i split the angle in half, and since that imaginary horizon is parrallel to the ground, it makes a right angle (90 degrees) with the distance to the ground (overhead point to "you" point) so that is why I did 90-half the cone angle. it made sense to me when i drew it out, but i'm not sure if i am explaining it correctly or if that is just totally incorrect.

    so i guess i am still :confused:
  18. Apr 14, 2006 #17


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  19. Apr 14, 2006 #18
    ohhh i did not know that, thanks! so does this look right, then:
    tan(47.8)=x/55566m, so x=61.27km?
    it still says this is wrong, so am i setting it up wrong still?

  20. Apr 14, 2006 #19

    Doc Al

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    You're setting it up just fine, but check your arithmetic for the distance the plane flies.
  21. Apr 14, 2006 #20
    aha! so it is just off by one decimal place:cool:

    thanks everyone, i get it!
  22. Mar 5, 2008 #21
    Hey peoples this really just helped me with an assignment (same question : P) but I don't know if I have everything straight or not, so if anyone is willing to look through the whole thought process and tell me if something's wrong...

    v : be the speed of the plane
    s : be the speed of sound
    h : be the height of the plane from the ground
    t : be the time from when the plane is directly overhead of the stationary observer (time zero) to when the sonic boom is heard by the observer (t).

    So after constructing a diagram with the points "plane at time zero" (I), "t" (F), "the observer" (O), and the line from I to O... the idea is to construct two triangles, one from O to I to F and one from I to F to the point P where the straight line IP is perpendicular and intersecting the line FO. Angle(OIF) and Angle(IPF) are right angles (you should be able to see why).

    I-------------F (bad picture but hopefully you get the gist of it)
    |........./ P

    So from what I understand the idea is to find two expressions for sin(theta) where theta is the angle from IF to OF and to equate the two equations.

    The first equation is derived using sin(theta)=opposite/hypotenuse where opposite is s*t (length of IP) and hypotenuse is v*t (length of IF) such that

    sin(theta) = (s*t)/(v*t) = s/v = 1/M where M is the Mach number

    The second equation is derived using sin(theta) again where opposite is h (the height and length of IO) and the hypotenuse is sqrt(h^2 + (v*t)^2) such that

    sin(theta) = h/sqrt(h^2 + (v*t)^2) (Pythagorean's Theorem is implicit here)

    Equating the two:

    1/M = h/sqrt(h^2 + (v*t)^2) (isolating for h)

    h = sqrt[(v*t)^2/(M^2 - 1)]

    Does this look alright? Plugging in the initial values I get 6.127 km high (right answer?)
  23. May 12, 2008 #22
    Thanks for the wonderful explanation, Mastertinker! Now I finally understand how to solve this problem.
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