What Is the Relationship Between Shock Cones and Sonic Booms?

[motherlovebone]

Okay, I got this worksheet today and my teacher didn't explain to me what shock cones were. So here's the problems, and what I did so far to work them out.

"An aircraft makes a supersonic pass 382 m over an observer. the shock wave pounds the observer after the plane is 405 m past her position (line of sight). what was the aircraft's ground speed if the flight path was level and the air temperature was 16 degrees Celsius?"

I started out by doing the speed of sound using the temperature correction formula, 331 + 0.6T, to get 340.6 m/s as the speed of sound. But now I am stuck as to how to find the ground speed of the plane.

This one, I have no clue on, so any hints will help.
"A light cone of 30 degrees is observed in crown glass which allows light to travel at 2.0 x 10^8 m/s. How fast was the particle supposedly going? Then find the real maximum light cone angle for a relativistically allowed velocity in this glass."

 

[lightgrav]

tan(30) = c_material/v , so v = ? (this is not allowed to be greater than c)
... the speed of light in a material = c/n , you'll have to look up the index of refraction for crown glass (probably about 1.5)

 

[Astronuc]

Assume that the boom travels straight down from the plane in time t = elevation (382 m) / speed of sound. In that time, the plane has traveled 405 m.

I believe that the second question may pertain to Cerenkov radiation.
http://en.wikipedia.org/wiki/Cherenkov_effect
http://www.physics.upenn.edu/balloon/cerenkov_radiation.html

Try thinking about the definition of index of refraction in the glass.

 

[FredGarvin]

For the first one, the speed is related to the angle of incidence of the shockwave, and thus to the aircraft's groundspeed and altitude.

Imagine the aircraft directly over the person's head. Now also imagine that the shockwave trails the aircraft, creating a cone with an angle \alpha relative to the ground (and the aircraft by geometry). If you call the altitude z and the distance between the person and where the shockwave meets the ground x you can create a relationship

\alpha = arctan \frac{z}{x}

You also need the relationship that Ma = \frac{1}{sin \alpha}

If you combine the two equations, you get the result

Ma = \frac{1}{sin(arctan\frac{z}{Vt})}

You have all the information to now solve for V