Practicing Success
The angle of elevation of an Aeroplane from a point on the ground is 60°. After flying for 8.66 seconds, the elevation changes to 30°. If the Aeroplane is flying at a height of 3000 meters, then the speed of the Aeroplane in m/s is? |
456 500 400 100 |
400 |
Aeroplane fly from A to P, Now, In ΔABC tan 60° = \(\frac{AB}{BC }\) ⇒ \(\frac{\sqrt {3}}{1 }\) = \(\frac{AB}{BC }\) Now, In ΔPQC tan 30° = \(\frac{PQ}{QC }\) ⇒ \(\frac{1}{\sqrt {3}}\) = \(\frac{PQ}{QC }\) or ⇒ \(\frac{\sqrt {3}}{3}\) = \(\frac{PQ}{QC }\) Here PQ = AB = height = 3000m, So ⇒\(\sqrt {3}\)R = 3000 m ⇒1R = 1000 \(\sqrt {3}\) m Here, Distance traveled by aeroplane = QB ⇒ QB = QC - BC = 3 - 1 = 2R ⇒2R = 2000 \(\sqrt {3}\) m Therefore, Speed = \(\frac{2000\sqrt {3}}{8.66}\) = \(\frac{2000}{5}\) = 400 ms |