Question

The shock wave off the cockpit of the FA 18 in Figure has an angle of about$\mathbf{60}\mathbf{°}\mathbf{.}$The airplane was traveling at about$\mathbf{1350}\mathbf{\text{\hspace{0.17em}km}}\mathbf{/}\mathbf{\text{h}}$when the photograph was taken. Approximately what was the speed of sound at the airplane’s altitude?

Short answer

The speed of the sound at airplane’s altitude is$325\text{\hspace{0.17em}m}/\text{s}$

Step-by-step solution

The given data

1. Angle of shock wave off the cockpit is$\mathrm{\theta }={60}^0$
2. Speed of airplane is${\mathrm{V}}_{\mathrm{a}}=1350\text{\hspace{0.17em}km}/\text{hor}375\text{\hspace{0.17em}m}/\text{s}$

Understanding the concept of shock wave

We use the concept of the shock wave. First, we convert the given velocity of the airplane in m/s, and then, using the equation of shock wave, we find the speed of sound.

Formula:

The speed of the sound using equation of shock wave,

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

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

Calculation of the speed of sound at airplane’s altitude

Using the given values and equation (i), we can get the speed of the sound as:

$\begin{array}{c}\sin {60}^0=\frac{{\mathrm{V}}_{\mathrm{s}}}{375\text{\hspace{0.17em}m}/\text{s}}\\ {\mathrm{V}}_{\mathrm{s}}=\sin {60}^0\times 375\text{\hspace{0.17em}m}/\text{s}\\ =324.75\text{\hspace{0.17em}m}/\text{s}\\ \approx 325\text{\hspace{0.17em}m}/\text{s}\end{array}$

Hence, the value of the speed of the sound is $325\text{\hspace{0.17em}m}/\text{s}$.