© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the physical situation
A person on the ground hears the sound of a jet plane apparently coming from the north at an angle of 60° with the ground, but the plane is actually directly overhead by the time the sound reaches the person. We denote the speed of the plane by $u$ and the speed of sound by $v$.
Step 2: Set up the geometry of the problem
Let the horizontal distance of the plane from the person at the instant it emits sound be $x$, and the altitude of the plane be $h$. At that moment, the sound travels along the line connecting the plane’s position to the person. Therefore, the distance the sound travels is $\sqrt{x^2 + h^2}$.
Step 3: Express the angle condition
The plane is heard at a 60° angle from the horizontal. From the geometry of right triangles, if $\theta$ is the angle from the horizontal, then
\[
\tan\theta = \frac{\text{vertical distance}}{\text{horizontal distance}} = \frac{h}{x}.
\]
Given $\theta = 60^\circ$,
\[
\tan 60^\circ = \sqrt{3} = \frac{h}{x}.
\]
Thus,
\[
h = \sqrt{3}\,x.
\]
Step 4: Relate time of sound travel to the plane’s motion
The time $T$ taken by the sound to reach the person is
\[
T = \frac{\sqrt{x^2 + h^2}}{v}.
\]
During this same time $T$, the plane moves horizontally at speed $u$. Since the plane is directly overhead the person by the time the sound arrives, the horizontal distance the plane traveled in time $T$ must have been $x$. In other words:
\[
x = u \times T = u \times \frac{\sqrt{x^2 + h^2}}{v}.
\]
Hence,
\[
x = \frac{u}{v} \sqrt{x^2 + h^2}.
\]
Step 5: Substitute $h$ in terms of $x$ into the time relation
From Step 3, $h = \sqrt{3}\,x$. Substitute into $\sqrt{x^2 + h^2}$:
\[
\sqrt{x^2 + (\sqrt{3} x)^2} = \sqrt{x^2 + 3x^2} = \sqrt{4x^2} = 2x.
\]
Thus the time relation becomes
\[
x = \frac{u}{v} \cdot 2x \quad \Longrightarrow \quad 1 = 2 \frac{u}{v}.
\]
Therefore,
\[
\frac{u}{v} = \frac{1}{2} \quad \Longrightarrow \quad u = \frac{v}{2}.
\]
Step 6: Conclude the speed of the plane
The speed of the plane is therefore
\[
u = \frac{v}{2}.
\]
Final Answer
The correct speed of the plane is $\frac{v}{2}$.
