A jet plane passes over you at a height of $\mathbf{5000}\mathbf{\text{\hspace{0.17em}m}}$and a speed of Mach 1.5.

(a) Find the Mach cone angle (the sound speed is $\mathbf{331}\mathbf{}\mathbf{\text{\hspace{0.17em}m}}\mathbf{/}\mathbf{\text{s}}$).

(b) How long after the jet passes directly overhead does the shock wave reach you?

1. Height of the jet plane is$\mathrm{h}=5000\text{\hspace{0.17em}m}$.
2. Speed of Mach is${\mathrm{V}}_{\mathrm{s}}=1.5\mathrm{V}$

We use the concept of the shock wave. Using the equation of Mach cone angle, we can find that angle. To find the time, we find the distance where the Mach cone intersects the surface of the earth. We can find that distance from the height and the Mach angle. Using the equation of velocity, we can solve the time taken by the shock wave to reach us.

Formula:

The speed of the sound using equation of shock wave (or the Mach cone angle),

$\mathbf{sin\theta }\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{s}}}{{\mathbf{V}}_{\mathbf{a}}}$ …(i)

Where,${\mathbf{V}}_{\mathbf{a}}$is the speed of the airplane.

Plugging the values in equation (i), we can get the Mach cone angle as:

$\begin{array}{c}\mathrm{sin\theta }=\frac{343\text{\hspace{0.17em}m}/\text{s}}{1.5(343\text{\hspace{0.17em}m}/\text{s})}\\ =\frac{1}{1.5}\\ =0.66\\ \mathrm{\theta }={\sin }^{-1}0.66\\ ={42}^0\end{array}$

Hence, the value of the Mach cone angle is ${42}^0$.

In the following figure, P is the position of plane and O is the position of observer.

We consider that Mach cone intersects earth’s surface at distance d. We can find d using the angle and height.

$\mathrm{tan\theta }=\frac{\mathrm{h}}{\mathrm{d}}$

$\mathrm{d}=\frac{\mathrm{h}}{\mathrm{tan\theta }}$

The time taken by the plane to travel distance d is

$\mathrm{t}=\frac{\mathrm{d}}{{\mathrm{V}}_{\mathrm{s}}}$

Plugging the values of d and ${\mathrm{V}}_{\mathrm{s}}$, we get

$\mathrm{t}=\frac{\mathrm{h}}{{\mathrm{V}}_{\mathrm{s}}\mathrm{tan\theta }}$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{t}=\frac{\mathrm{h}}{{\mathrm{V}}_{\mathrm{s}}\mathrm{tan\theta }}\\ \mathrm{t}=\frac{5000\text{\hspace{0.17em}m}}{\left(1.5\right)(331\text{\hspace{0.17em}m}/\text{s})\tan {42}^0}\\ =11\text{\hspace{0.17em}s}\end{array}$

Hence, the value of the required time is $11\text{\hspace{0.17em}s}$.