Step-by-step Solution:

Correct Answer: B (8000/3m)

This is a classic heights and distances problem that can be solved by setting up two right-angled triangles and using trigonometric ratios.

Step 1: Visualize and Set Up the Problem

Let's define the points involved:

• Let O be the observation point on the ground.
• Let A be the point on the ground directly below the two airplanes.
• Let P1 be the higher airplane at a height of 4000m ($AP_1 = 4000$).
• Let P2 be the lower airplane at a height of $h_2$ ($AP_2 = h_2$).
• Let x be the horizontal distance from the observer to the airplanes ($OA = x$).

This setup forms two right-angled triangles: $\triangle OAP_1$ and $\triangle OAP_2$. The angles of elevation are given as:

• $\angle AOP_1 = 60^\circ$
• $\angle AOP_2 = 30^\circ$

Our goal is to find the vertical distance between the airplanes, which is $P_1P_2 = AP_1 - AP_2 = 4000 - h_2$.

Step 2: Apply Trigonometric Ratios

We can use the tangent function ($\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}}$) for both triangles.

For the higher airplane ($\triangle OAP_1$):

$$ \tan(60^\circ) = \frac{AP_1}{OA} = \frac{4000}{x} $$

Since $\tan(60^\circ) = \sqrt{3}$, we have:

$$ \sqrt{3} = \frac{4000}{x} \implies x = \frac{4000}{\sqrt{3}} $$

For the lower airplane ($\triangle OAP_2$):

$$ \tan(30^\circ) = \frac{AP_2}{OA} = \frac{h_2}{x} $$

Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, we have:

$$ \frac{1}{\sqrt{3}} = \frac{h_2}{x} \implies h_2 = \frac{x}{\sqrt{3}} $$

Step 3: Calculate the Height of the Lower Airplane ($h_2$)

Now we substitute the expression for $x$ from the first equation into the second equation:

$$ h_2 = \frac{1}{\sqrt{3}} \left( \frac{4000}{\sqrt{3}} \right) = \frac{4000}{3} $$

The height of the lower airplane is $\frac{4000}{3}$ meters.

Step 4: Find the Vertical Distance Between the Airplanes

The required distance is the difference between their heights:

Distance $= h_1 - h_2 = 4000 - \frac{4000}{3}$

Distance $= \frac{12000 - 4000}{3} = \frac{8000}{3}$ meters.