A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angl

Q: Why is the plane vertically above the person when the sound is heard?

The problem states that at the exact moment the person hears the sound (which was emitted earlier), they look up and see the plane directly overhead. This means the plane has traveled the horizontal distance exactly in the time the sound traveled the diagonal distance.

Q: Why do we use $\cos 60^\circ$ and not $\sin 60^\circ$?

The angle $60^\circ$ is given with the horizontal (ground level). In our right-angled triangle, the horizontal distance traveled by the plane is the adjacent side (base) to this angle, and the path of the sound is the hypotenuse. The ratio of base to hypotenuse is the cosine function.

Q: Does the speed of the plane affect the speed of the sound it emits?

No! The speed of sound in a medium (like air) depends only on the properties of the medium, not on the velocity of the source. The sound travels at speed $v$ regardless of how fast the plane is moving.

Q: What if the plane was flying at the speed of sound ($v_p = v$)?

If $v_p = v$, then $\cos \theta = 1$, which means $\theta = 0^\circ$. The plane would be right at the horizon, and you would hear the sound exactly as it passes you, creating a sonic boom.

Q: Is this related to the Doppler effect?

While both involve moving sources of sound, this specific problem is purely about the kinematics of wave propagation (distance = speed $\times$ time). The Doppler effect deals with the *change in frequency* of the sound, which isn't asked here.

Question

A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle $60^{\circ}$ with ground level. But he finds the aeroplane right vertically above his position. If $v$ is the speed of sound, then speed of the plane is

Options:

(A) $\frac{\sqrt{3}}{2}v$

(B) $v$

(C) $\frac{2v}{\sqrt{3}}$

(D) $\frac{v}{2}$

Sigma Percentile · Accepted Answer

Have you ever looked up at the sky after hearing a jet plane, only to find that the plane is far ahead of where the sound seems to be coming from? This classic phenomenon happens because sound takes a noticeable amount of time to travel from the high-altitude plane to your ears.

### Analyzing the Setup

Imagine you are standing on the ground at point $A$. A jet plane is flying horizontally high above you. At a specific instant, let's call it $t = 0$, the plane is at position $P_1$. At this exact moment, the plane emits a sound wave that begins its journey towards you.

**Here is the crucial part:** The sound doesn't reach you instantly. It travels along the diagonal path $P_1A$ at the speed of sound, $v$. By the time this sound wave finally reaches your ears at time $t$, the plane has continued flying forward.

According to the problem, when you hear the sound, you look straight up and see the plane directly overhead at a new position, $P_2$. This means the plane has traveled the horizontal distance $P_1P_2$ in the exact same time $t$ that the sound took to travel the diagonal distance $P_1A$.

### The Master Equation

Let's quantify these distances. If the speed of the plane is $v_p$, the horizontal distance it covers is:

$$P_1P_2 = v_p 	imes t$$

Meanwhile, the sound travels at speed $v$, so the diagonal distance it covers is:

$$P_1A = v 	imes t$$

Now, let's look at the geometry. The path of the plane, the path of the sound, and the vertical line from the plane's new position to you form a right-angled triangle, $	riangle P_1P_2A$. The angle between the plane's horizontal path and the sound's path is given as $60^\circ$.

In this right-angled triangle, we can use basic trigonometry. The cosine of an angle is the ratio of the adjacent side (the base) to the hypotenuse.

$$\cos 60^\circ = \frac{P_1P_2}{P_1A}$$

### Final Calculation

Now, we simply substitute the distances we found into our trigonometric equation:

$$\cos 60^\circ = \frac{v_p 	imes t}{v 	imes t}$$

The time $t$ cancels out beautifully from the numerator and the denominator. We also know that $\cos 60^\circ = \frac{1}{2}$.

$$\frac{1}{2} = \frac{v_p}{v}$$

Rearranging this equation to solve for the speed of the plane, we get:

$$v_p = \frac{v}{2}$$

**The speed of the plane is exactly half the speed of sound.** This elegant result shows how simple geometry and kinematics can unravel the everyday mysteries of the world around us!

Animated Solution for Physics - Waves: A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle $6 0^{\circ}$ with ground level. But he finds the aeroplane right vertically above his position. If $v$ is the speed of sound, then speed of the plane is

Select Answer:

Visualized Solution

$Visualizing the Scenario$

$Simultaneous Motion$

$Distance Equations$

$Trigonometric Relation$

$Substituting Values$

$Final Velocity$

$The Way Forward$

The Sigma Insight: Wave Equation and Wave Speed

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

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

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

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

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

.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 Velocity

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