A hot air balloon is rising vertically upwards at a constant...

Let the distance of the observer from the launch pad be $d$ and the uniform speed of the balloon be $v$.

Let the heights of the balloon at the three instants be $h_1$, $h_2$, and $h_3$.

From the given angles of elevation:
$h_1 = d \tan 30^\circ = \dfrac{d}{\sqrt{3}}$
$h_2 = d \tan 45^\circ = d$
$h_3 = d \tan 60^\circ = d\sqrt{3}$

The distance covered by the balloon in the first $10$ minutes is:
$h_2 - h_1 = d - \dfrac{d}{\sqrt{3}} = d\left(\dfrac{\sqrt{3}-1}{\sqrt{3}}\right)$

Since the speed is uniform, $v = \dfrac{h_2 - h_1}{10} = \dfrac{d(\sqrt{3}-1)}{10\sqrt{3}}$.

The further distance to be covered for the angle of elevation to become $60^\circ$ is:
$h_3 - h_2 = d\sqrt{3} - d = d(\sqrt{3}-1)$

The time taken to cover this further distance is:
$t = \dfrac{h_3 - h_2}{v} = \dfrac{d(\sqrt{3}-1)}{\dfrac{d(\sqrt{3}-1)}{10\sqrt{3}}} = 10\sqrt{3}$ minutes.

Answer: $10\sqrt{3}$