1. System Visualization

The core concept is the time lag caused by the finite speed of sound. The aircraft is at one position ($P_1$) when it emits the sound, but by the time that sound reaches the observer, the aircraft has moved to a new position ($P_2$).

2. Geometric Relations

Using the altitude $h$ and the elevation angles, we can define the horizontal distances from the observer’s vertical axis:

• Sound Position ($P_1$): Horizontal distance $x_1 = h \cot \beta$.
• Visual Position ($P_2$): Horizontal distance $x_2 = h \cot \alpha$.
• Distance Traveled: The plane flew the distance $\Delta x = x_1 – x_2$ while the sound was traveling.

3. Kinematics Equation

We equate the time taken for two events:

1. Sound Travel Time: The sound traveled the hypotenuse distance from $P_1$ to $O$. $$ t = \frac{\text{Distance}_{P_1 \to O}}{v_{sound}} = \frac{h \csc \beta}{v_s} $$
2. Flight Time: The plane flew from $P_1$ to $P_2$ in the exact same time $t$. $$ \Delta x = v_{plane} \cdot t $$

Combining these:

4. Relating to Angular Velocity

The problem gives the angular velocity $\omega$ when the plane is overhead. For an object moving in a straight line at altitude $h$, the relationship is:

$$ v = \omega h $$

Substituting this into our velocity ratio equation:

5. Final Calculation

Given Values: $\alpha = 53^\circ$, $\beta = 37^\circ$, $v_s = 330$ m/s, $\omega = \frac{1}{8}$ rad/s.

Note: $\sin 37^\circ = 0.6$, $\cot 37^\circ = 1.33$, $\cot 53^\circ = 0.75$.

$$ h = \frac{330 \cdot 0.6}{0.125} \left( \frac{4}{3} – \frac{3}{4} \right) $$ $$ h = \frac{198}{1/8} \left( \frac{7}{12} \right) $$ $$ h = 198 \cdot 8 \cdot \frac{7}{12} = 198 \cdot \frac{56}{12} $$