Best Answer k-man, 21 February 2013 - 09:28 PM

I'm making a couple of simplifying assumptions here:

1) we're ignoring the curvature of the earth and the triangle KJM with sides 10000ft, 15000ft, 6000ft is planar.

2) the sonic boom is a momentary event that occurred and the sound from this event was moving with a constant speed of 1100ft/sec in all directions.

Let's define our coordinates in the following way (1/100 ft scale):

Kiru (K) is in the origin (0,0,0)

John (J) is 15,000 ft along the X axis (150,0,0)

Michelle (M) is on the XY plane. We can calculate her coordinates from the following 2 equations:

1) x^{2}+y^{2}=100^{2}

2) (x-150)^{2}+y^{2}=60^{2}

Solving these equations gives us the coordinates for M (289/3, sqrt(6479)/3, 0) or approximately (96.33, 26.33, 0)

Let **r **be the distance from the boom to the origin (K), then the distance to M is r+22 and the distance to J is r+77. We can now construct 3 equations describing 3 spheres centered in K, M and J and corresponding radii of r, r+22 and r+77.

1) x^{2}+y^{2}+z^{2}=r^{2}

2) (x-150)^{2}+y^{2}+z^{2}=(r+77)^{2}

3) (x-289/3)^{2}+(y-sqrt(6479)/3)^{2}+z^{2}=(r+22)^{2}

Solving these equations we get

x = (16571 - 154 r)/300

y = (24706 r - 506819) / (300 sqrt(6479))

z = +/- sqrt(-50259 r^{2}+ 16141917 r - 565551724) / (5 sqrt(6479))

Now we need to find the range of r for which there are real solutions for z: (-50259 r^{2 }+ 16141917 r - 565551724) > 0

This gives us a valid range for r:

(16141917-5 sqrt(5874609121953))/100518 < r < (16141917+5 sqrt(5874609121953))/100518, or approximately 40.0239 < r < 281.151

This range corresponds to a range of solutions for where a jet can be. If you know how high the jet was flying (z) then you can figure out the rest using the formulas above.