An aircraft is flying at a level height in a straight line. When you see it at an elevation alpha =

Q: Why do we use the same time $t$ for both the sound and the aircraft?

Because the event of the sound reaching your ears happens exactly when you see the aircraft at $53^\circ$. The sound started its journey when the aircraft was at $37^\circ$. So, the duration of the sound's journey is identical to the duration of the aircraft's flight between those two points.

Q: Why is the distance traveled by sound $h / \sin \beta$?

If you draw a right-angled triangle with the observer, the point on the ground directly below the aircraft's initial position, and the aircraft itself, the height is $h$ and the angle of elevation is $\beta$. The hypotenuse, which is the path of the sound, is $h / \sin \beta$.

Q: How did we get the distance the aircraft traveled as $h(\cot \beta - \cot \alpha)$?

The horizontal distance from the observer to the aircraft at elevation $\beta$ is $h \cot \beta$. At elevation $\alpha$, it is $h \cot \alpha$. The distance the aircraft flew is simply the difference between these two horizontal positions.

Q: Why is $\omega = v/h$ when the aircraft is overhead?

Angular velocity $\omega$ is defined as the perpendicular component of velocity divided by the distance to the observer. When the aircraft is directly overhead, its entire velocity $v$ is perpendicular to the vertical line of sight, and the distance is exactly the height $h$.

Q: What if the aircraft was flying away from the observer?

If it were flying away, the elevation angle would decrease over time. Since the problem states you hear the sound from $37^\circ$ when you see it at $53^\circ$, it must be moving towards you (from a lower elevation angle to a higher one).

Question

An aircraft is flying at a level height in a straight line. When you see it at an elevation $\alpha = 53^\circ$ above the horizontal, you hear its sound coming from an elevation $\beta = 37^\circ$ above the horizontal. When the aircraft passes a location vertically above your head, its angular velocity relative to you is $\omega = 0.125 	ext{ rad/s}$. Speed of sound in air is $v_s = 330 	ext{ m/s}$. If transit time of light from the aircraft to you is negligible as compared to that of the sound, calculate altitude of the aircraft.

Sigma Percentile · Accepted Answer

Have you ever looked up at the sky, spotted a high-flying jet, and noticed that the roar of its engines seems to trail far behind it? This everyday phenomenon is a beautiful demonstration of the finite speed of sound, and it's exactly what this problem asks us to decode.

Let's embark on this thrilling chase to find the aircraft's altitude!

### Visualizing the Delay

Imagine standing at point $O$ on the ground. You look up at an angle of $\alpha = 53^\circ$ and see the aircraft at position $P_2$.

However, the sound hitting your ears right now isn't from $P_2$. It was emitted earlier, when the aircraft was further back at position $P_1$, corresponding to an elevation of $\beta = 37^\circ$.

Because light travels almost instantaneously, you see the plane at $P_2$ exactly when the sound from $P_1$ finally reaches you. This means the time $t$ it took for the sound to travel from $P_1$ to your ears is exactly the same time the aircraft took to fly from $P_1$ to $P_2$.

### The Kinematics of the Chase

Let's translate this physical reality into mathematics.

The distance the sound traveled is $OP_1 = v_s t$, where $v_s$ is the speed of sound.

Using the right-angled triangle formed by the aircraft's height $h$ and the line of sight $OP_1$, we can express this distance as $OP_1 = \frac{h}{\sin\beta}$.

Therefore, the time elapsed is:
$$t = \frac{h}{v_s \sin\beta}$$

During this exact same time $t$, the aircraft, flying at a constant velocity $v$, covered a horizontal distance $P_1 P_2 = v t$.

We can also find this distance geometrically. The horizontal position of $P_1$ is $h \cot\beta$, and the horizontal position of $P_2$ is $h \cot\alpha$. The distance flown is simply the difference:
$$P_1 P_2 = h \cot\beta - h \cot\alpha$$

Equating our two expressions for the aircraft's distance, we get our master equation:
$$v \left( \frac{h}{v_s \sin\beta} ight) = h (\cot\beta - \cot\alpha)$$

Notice how the unknown height $h$ beautifully cancels out here! This allows us to find the aircraft's velocity:
$$v = v_s \sin\beta (\cot\beta - \cot\alpha)$$

### The Overhead Clue

We are given one more crucial piece of the puzzle: when the aircraft is directly overhead, its angular velocity is $\omega = 0.125 	ext{ rad/s}$.

**Angular velocity** is defined as the component of velocity perpendicular to the line of sight, divided by the distance to the observer.

When the plane is directly overhead, its entire velocity $v$ is perpendicular to your vertical line of sight, and its distance from you is exactly its altitude $h$.

Therefore, we can write:
$$\omega = \frac{v}{h} \implies v = \omega h$$

### Bringing It All Together

Now, we equate our two expressions for the velocity $v$:
$$\omega h = v_s \sin\beta (\cot\beta - \cot\alpha)$$

Isolating the altitude $h$, we get:
$$h = \frac{v_s \sin\beta}{\omega} (\cot\beta - \cot\alpha)$$

All that's left is to substitute the given values. We know $v_s = 330 	ext{ m/s}$ and $\omega = 1/8 	ext{ rad/s}$.

Using the classic $3-4-5$ triangle properties for our angles, we have $\sin 37^\circ = 3/5$, $\cot 37^\circ = 4/3$, and $\cot 53^\circ = 3/4$.

Plugging these in:
$$h = \frac{330 	imes (3/5)}{1/8} \left( \frac{4}{3} - \frac{3}{4} ight)$$

$$h = 198 	imes 8 	imes \left( \frac{16 - 9}{12} ight)$$

$$h = 1584 	imes \frac{7}{12}$$

$$h = 132 	imes 7 = 924 	ext{ m}$$

And there we have it! By carefully tracking the delayed arrival of sound, we've successfully deduced that the aircraft is flying at an altitude of **$924 	ext{ meters}$**.

Enter Numerical Value:

Visualized Solution

$Understanding the Physical Situation$

$Kinematics of Sound and Aircraft$

$Equating the Time Intervals$

$Expressing Aircraft Velocity$

$Angular Velocity Overhead$

$Solving for Height$

$Substituting Values$

$Final Calculation$

$Extensions and Reflections$

The Sigma Insight: Relative Velocity

Visualizing the Delay

The Kinematics of the Chase

The Overhead Clue

Bringing It All Together

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Understanding the Physical Situation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kinematics of Sound and Aircraft

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating the Time Intervals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Expressing Aircraft Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Angular Velocity Overhead

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Height

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Extensions and Reflections