Question

A supersonic jet traveling at Mach 2.10 at an altitude of h 18,000 m is directly over a person at timet-o as shown in the figure below. Assume the average speed of sound in air is 335 m/s over the path of the sound. Observer hears the "boom" Observer (a) At what time will the person encounter the shock wave due to the sound emitted at t02 (b) Where will the plane be when this shock wave is heard?

Answer 1

a) The altitude at which the flight is flying is given as 18000m. The time taken for the sound to travel this distance of 18000 m is given by the equation $t=\frac{s}{v}$ where $v$ is the velocity of sound in air and $s$ is the distance between the flight and the observer, ie 18000 m.

$t=53.73 s$

b) the plane is travelling at Mach 2.10. that means its travelling 2.10 times faster than the speed of sound in undisturbed air. the speed of sound in undisturbed air is 332m/s. this would mean that the flight travelled $s=53.73 *332=17838.36m$ away from the original position.

if we take a right angled triangle, with one of the sides as 18000 m and other as 17838.36, then we can find the indicated angle.

when we take the tangent of the given angle, as given from the figure, $tan\theta =1.0090$

$\theta =45.2567^o$

this is an exact calculation. Instead, if we substitute Mach 2.10 velocity value as 335 m/s, the horizontal distance travelled by the plane would come to about 17999.5 m and when we find the angle, it would come to 45 degrees. so for calculation purpuse, we can take Mach velocity as 335, but officially, Mach 1 value would mean 332m/s.