🎥 Video solution / Text Solution of this question is given below:
Let the higher airplane be at a height of $4000\,\text{m}$.
From a point on the ground, the angles of elevation to the two airplanes are
$60^\circ$ (upper plane) and $30^\circ$ (lower plane).
Let the horizontal distance from the observer to the airplanes be $x$.
For the upper airplane:
$\tan 60^\circ = \dfrac{4000}{x}$
$\sqrt{3} = \dfrac{4000}{x}$
$\Rightarrow x = \dfrac{4000}{\sqrt{3}}$.
For the lower airplane with height $h$:
$\tan 30^\circ = \dfrac{h}{x}$
$\dfrac{1}{\sqrt{3}} = \dfrac{h}{4000/\sqrt{3}}$
Thus,
$h = \dfrac{4000}{3}$.
Now the vertical distance between the two airplanes:
$4000 - \dfrac{4000}{3}
= \dfrac{8000}{3}.$
Final Answer: $\displaystyle \frac{8000}{3}\text{ m}$.