A supersonic aircraft flies at an altitude of 1.8 km where temperature is 4°C. Determine the speed of the aircraft if its sound is heard 4 seconds after its passage over the head of an observer. Take R = 287 J/kg K and γ = 1.4.

Solution:

Altitude of the aircraft, $ = 1.8 km = 1800 m $

Temperature, $ T = 4 + 273 = 277 K $

Time, $ t = 4 s $

Speed of the aircraft V :

Refer Let, O represent the observer and A the position of the aircraft just vertically over the observer. After 4 seconds, the aircraft reaches the position represented by the point B. Line AB represents the wave front and $\alpha$ the Mach angle.

From Fig., we have,

$ \tan \alpha=\frac{1800}{4 V}=\frac{450}{V}...(1) \\ $

But, Mach number,

$ M=\frac{C}{V}=\frac{1}{\sin \alpha} \\ $

$ V=\frac{C}{\sin \alpha}....(2) \\ $

Substituting the value of $V$ in eqn. (i), we get, $$ \tan \alpha=\frac{450}{(C / \sin \alpha)}=\frac{450 \sin \alpha}{C} \ $$

$$ \frac{\sin \alpha}{\cos \alpha}=\frac{450 \sin \alpha}{C} \ or \cos \alpha=\frac{C}{450}....(3) \\ $$

But $C=\sqrt{\gamma R T}$, where C is the sonic velocity.

$ R=287 \mathrm{~J} / \mathrm{kg} \mathrm{K} \\ $ and $ \gamma=1.4...(Given) \\ $

$ \therefore \quad C=\sqrt{1.4 \times 287 \times 277}=333.6 \mathrm{~m} / \mathrm{s} \\ $

Substituting the value of $C$ in eqn. (ii), we get, $$ \cos \alpha=\frac{333.6}{450}=0.7413 \ $$

$$ \therefore \quad \sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\sqrt{1-0.7413^{2}}=0.6712 \\ $$

Substituting the value of $\sin \alpha$ in eqn. (ii), we get,

$$ V=\frac{C}{\sin \alpha}=\frac{333.6}{0.6712}=497 \mathrm{~m} / \mathrm{s}=\frac{497 \times 3600}{1000}=\mathbf{1 7 8 9 . 2} \mathbf{~ k m} / \mathbf{h} \text { (Ans.) } \\ $$