Question

A supersonic jet traveling at Mach 2.0 at an altitude of 9500 m passes directly over an observer on the ground. Where will the plane be relative to the observer when the latter hears the sonic boom? (See Fig. 12-39.)

Short answer

The plane is at a distance of \(16454.5\;{\rm{m}}\) at which the observer hears the sonic boom.

Step-by-step solution

Determination of angle of shock cone

The sine component of the angle of the shock cone is equal to the inverse of the Mach number of the system. The angle of the shock cone represents the displacement of the system with respect to sound from the observer.

Given information

Given data:

The Mach number for jet is \({\rm{Mach number}} = 2.0\).

The height of the jet from the ground is \(h = 9500\;{\rm{m}}\).

Evaluate the angle of shock cone and distance of jet relative to the observer

The diagrammatical representation for motion of jet is shown below:

The expression for the Mach number is given as:

\(\begin{aligned}{c}{\rm{Mach}}\;{\rm{number}} &= \frac{u}{c}\\\frac{c}{u} &= \frac{1}{{{\rm{Mach}}\;{\rm{number}}}}\end{aligned}\)

The angle of shock cone is calculated below:

\(\begin{aligned}{c}\theta &= {\sin ^{ - 1}}\left( {\frac{c}{u}} \right)\\ &= {\sin ^{ - 1}}\left( {\frac{1}{{{\rm{Mach number}}}}} \right)\\ &= {\sin ^{ - 1}}\left( {\frac{1}{{{\rm{2}}{\rm{.0}}}}} \right)\\ = 30^\circ \end{aligned}\)

The distance of jet relative to observer is calculated below:

\(\begin{aligned}{c}\tan \theta &= \frac{h}{d}\\d &= \frac{h}{{\tan \theta }}\\ &= \frac{{\left( {9500\;{\rm{m}}} \right)}}{{\tan \left( {30^\circ } \right)}}\\ &= 16454.5\;{\rm{m}}\end{aligned}\)

Hence, the plane is at a distance of \(16454.5\;{\rm{m}}\) at which the observer hears the sonic boom.