Sonic boom plane altitude problem

In summary: It should be 5556.6 m = 5.5566 km.In summary, the altitude of the jet fighter can be calculated by finding the angle of the sonic cone, dividing it by 2 to get the half angle, and then using the tangent function to find the distance from the plane to the point where the sonic boom is heard. This distance can then be added to the height of the observer to get the altitude of the plane.
  • #1
kellyneedshelp
41
0
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!
 
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  • #2
Why can't you simply use;

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

-Hoot
 
  • #3
sorry, I am not familiar with how to use that kind of equation, could you explain what it means, please? thanks!
 
  • #4
I meant, why does the velocity of the aircraft matter in how long it takes the sound to reach the ground?

-Hoot
 
  • #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!
 
  • #6
kellyneedshelp said:
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?

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.

-Hoot
 
  • #7
ok thank you, although apparently 4.116 km is not right either so i am still confused about this one...
 
  • #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
 
  • #9
so any ideas on how to factor in the speed of the aircraft with the speed at which the sound reaches the observer?
 
  • #10
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:
  • #11
Doc Al said:
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 where the plane is at the moment you hear the boom.

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:

-Hoot
 
  • #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...

thanks!
 
  • #13
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?
 
  • #14
anyone know if i am picturing the triangle correctly?
thanks!
 
  • #15
kellyneedshelp said:
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)
Why did you divide by 2?
 
  • #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
FredGarvin said:
Take a look here:
http://www.grc.nasa.gov/WWW/K-12/airplane/machang.html

You are not calculating the included angle, you are already calculating the half angle.

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?

thanks!
 
  • #19
kellyneedshelp said:
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?
You're setting it up just fine, but check your arithmetic for the distance the plane flies.
 
  • #20
aha! so it is just off by one decimal place:cool:

thanks everyone, i get it!
 
  • #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...

Let:
--------------------------------------------
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
|.../
|.../
|.../
|./
O

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?)
 
  • #22
Thanks for the wonderful explanation, Mastertinker! Now I finally understand how to solve this problem.
 

1. What is a sonic boom plane altitude problem?

A sonic boom plane altitude problem occurs when an aircraft traveling at supersonic speeds creates a loud noise or shockwave, known as a "sonic boom". This can be disruptive and potentially damaging to structures and the environment.

2. How does the altitude of the plane affect the sonic boom?

The altitude of the plane is a key factor in the intensity and distance of the sonic boom. The higher the altitude, the weaker and less disruptive the sonic boom will be as it dissipates over distance. However, if the plane is too low, the sonic boom can be stronger and more damaging.

3. What regulations are in place for supersonic flight to prevent sonic boom problems?

In the United States, the Federal Aviation Administration (FAA) has established regulations that prohibit supersonic flight over land. This is to prevent sonic boom disturbances to communities and structures. However, there are exceptions for military and research purposes.

4. Can technology or design changes reduce the sonic boom problem?

There have been advancements in technology and design that aim to reduce the intensity of sonic booms. For example, NASA has been researching and testing a concept called "low boom" aircraft, which would produce a softer and less disruptive sonic boom. However, it is still a work in progress and has not been implemented on a large scale yet.

5. Are there any health risks associated with sonic booms?

There is currently no evidence to suggest that sonic booms pose any direct health risks to humans. However, the loud noise can be startling and potentially harmful to those with sensitive hearing or those who are exposed to it frequently. It can also cause damage to structures and disrupt wildlife.

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