Q. The angle of elevation of a jet fighter from a point $ A $ on the ground is $ {{60}^{o}} $ . After a flight of $ 10\,\,s $ , the angle of elevation changes to $ {{30}^{o}} $ . If the jet is flying at a speed of $ 432\,\,km/h $ . Find the constant height at which the jet is flying.
Jharkhand CECEJharkhand CECE 2009
Solution:
Since, speed of a flight be $ 432\,\,km/h $ .
$ \therefore $ Distance cover from $ A $ to $ B $ ,
$ d=432\times \frac{5}{18}\times 10=1200\,\,m $
Now, in $ \Delta CBD $ ,
$ \tan {{60}^{o}}=\frac{h}{x}\Rightarrow x=\frac{h}{\sqrt{3}} $ and in
$ \Delta CAD $ , $ \tan {{30}^{o}}=\frac{h}{d+x} $
$ \Rightarrow $ $ \frac{1}{\sqrt{3}}=\frac{h}{1200+\frac{h}{\sqrt{3}}} $
$ \Rightarrow $ $ 1200=\sqrt{3}h-\frac{h}{\sqrt{3}} $
$ \Rightarrow $ $ h=600\sqrt{3}m $
