91Ó°ÊÓ

1. Speed of plane is${\mathrm{V}}_{\mathrm{p}}=1.25\mathrm{V}$
2. Time of sonic boom to reach man at ground is$t=1.00\text{″¾¾±²Ô}$.
3. Speed of sound is $\mathrm{V}=330\text{″¾}/\text{s}$.

We use the concept of the shock wave. Using trigonometry, we can write the equation for height, and we can write distance in terms of the velocity of the plane. We find the angle from the shock wave equation, and then plugging it in the equation of height; we find the altitude of the plane.

Formula:

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

$\mathbf{²õ¾\pm ²Ôθ}\mathbf{=}\frac{\mathbf{V}}{{\mathbf{V}}_{\mathbf{p}}}$ …(¾±)

The tangent angle of a triangle,

$\mathbf{tan}\mathit{θ}\mathbf{=}\frac{\mathbf{y}}{\mathbf{x}}$ …(¾±¾±)

From the above figure, using equation (ii), we can write the height of the plane as:

$\begin{array}{c}\mathrm{³Ù²¹²Ôθ}=\frac{\mathrm{h}}{\mathrm{d}}\\ \mathrm{h}=\mathrm{d³Ù²¹²Ôθ}\end{array}$

The distance can be given as,

$\mathrm{d}={\mathrm{V}}_{\mathrm{p}}\mathrm{t}$

Now,

$\mathrm{h}=\left({\mathrm{V}}_{\mathrm{p}}\mathrm{t}\right)\mathrm{³Ù²¹²Ôθ}$

We know ${\mathrm{V}}_{\mathrm{p}}=1.25\mathrm{V}$.

$\mathrm{h}=\left(1.25\mathrm{Vt}\right)\mathrm{³Ù²¹²Ôθ}$ …(²¹)

Using shock wave equation (i), we get the Mach cone angle as:

$\begin{array}{c}\mathrm{²õ¾\pm ²Ôθ}=\frac{\mathrm{V}}{1.25\mathrm{V}}\\ =0.8\\ \mathrm{θ}={\sin }^{−1}\left(0.8\right)\\ ={53.1}^0\end{array}$

Hence, substituting the given values and the Mach angle in equation (a), we get the altitude as:

$\begin{array}{c}\mathrm{h}=(1.25×(330\text{″¾}/\text{s})×\left(60\text{ s}\right)×\tan \left({53.1}^0\right))\\ =3.30×{10}^4\text{″¾}\end{array}$

Hence, the value of altitude of the plane is, $3.30×{10}^4\text{″¾}$.