A hot air balloon is rising vertically at a constant speed of 5 m/s...

Let the initial height of the balloon be $h$ and the horizontal distance of the stone from the point directly below the balloon be $d$.

From the initial observation, the angle of depression is $30^\circ$. Thus,
$\tan 30^\circ = \dfrac{h}{d} \Rightarrow h = \dfrac{d}{\sqrt{3}}$

The balloon rises at a constant speed of $5 \text{ m/s}$ for $20$ seconds. The vertical distance covered is:
$5 \times 20 = 100 \text{ m}$

The new height of the balloon is $h + 100$. The new angle of depression is $60^\circ$. Thus,
$\tan 60^\circ = \dfrac{h + 100}{d} \Rightarrow h + 100 = d\sqrt{3}$

Substituting the value of $h$ from the first equation:
$\dfrac{d}{\sqrt{3}} + 100 = d\sqrt{3}$

$100 = d\sqrt{3} - \dfrac{d}{\sqrt{3}} = d\left(\dfrac{3 - 1}{\sqrt{3}}\right) = \dfrac{2d}{\sqrt{3}}$

$d = \dfrac{100\sqrt{3}}{2} = 50\sqrt{3} \text{ m}$

Answer: $50\sqrt{3} \text{ m}$