MAT Exam 2025 Paper

Concept:Use $\tan$ of the angle of elevation to find the horizontal distances of the aeroplane from the observing point.Explanation:Height of the aeroplane is $h = 3200\sqrt{3} \ \mathrm{m}$.After 25 seconds, the angle of elevation is $30^\circ$.Let the horizontal distance at that moment be $x$.$$\tan 30^\circ = \frac{h}{x}$$$$\frac{1}{\sqrt{3}} = \frac{3200\sqrt{3}}{x}$$$$x = 3200\sqrt{3} \times \sqrt{3} = 9600 \ \mathrm{m}$$Initially, the angle of elevation was $60^\circ$.Let the initial horizontal distance be $y$.$$\tan 60^\circ = \frac{h}{y}$$$$\sqrt{3} = \frac{3200\sqrt{3}}{y}$$$$y = 3200 \ \mathrm{m}$$Distance travelled in 25 seconds:$$d = x - y = 9600 - 3200 = 6400 \ \mathrm{m}$$Speed of the aeroplane:$$\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{6400}{25} = 256 \ \mathrm{m/s}$$Answer:The speed of the aeroplane is $256 \ \mathrm{m/s}$, so the correct option is A.