Let the height of flying of the aero-plane be $\mathrm{PQ}=\mathrm{BC}$ and point $\mathrm{A}$ be the point of observation.

$\mathrm{PQ}=\mathrm{BC}=2500 \mathrm{~m}, \angle \mathrm{PAQ}=45^{\circ}$ and $\angle \mathrm{BAC}=30^{\circ}$

In $\triangle \mathrm{PAQ}$,

$\tan 45^{\circ}=\frac{\mathrm{PQ}}{\mathrm{AQ}}$

$\Rightarrow 1=\frac{2500}{\mathrm{AQ}}$

$\Rightarrow \mathrm{AQ}=2500 \mathrm{~m}$

=> in $\triangle \mathrm{ABC}$,

$\tan 30^{\circ}=\frac{\mathrm{BC}}{\mathrm{AC}}$

$\begin{aligned}
&\Rightarrow \frac{1}{\sqrt{3}}=\frac{2500}{\mathrm{AC}} \\
&\Rightarrow \mathrm{AC}=2500 \sqrt{3} \mathrm{~m} \\
&\text { Now, } Q C=\mathrm{AC}-\mathrm{AQ} \\
&=2500 \sqrt{3}-2500 \\
&=2500(\sqrt{3}-1) \mathrm{m} \\
&=2500(1.732-1) \\
&=2500(0.732) \\
&=1830 \mathrm{~m} \\
&\Rightarrow \mathrm{PB}=\mathrm{QC}=1830 \mathrm{~m}
\end{aligned}
$

=>  the speed of the aero-plane $=\frac{P B}{15}$

$=\frac{1830}{15}$

$=122 \mathrm{~m} / \mathrm{s}$

$=122 \times \frac{3600}{1000} \mathrm{~km} / \mathrm{h}$

$=439.2 \mathrm{~km} / \mathrm{h}$

=> the speed of the aero-plane is $122 \mathrm{~m} / \mathrm{s}$ or $439.2 \mathrm{~km} / \mathrm{h}$.