The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30°. If the jet plane is flying at a constant height, then its height is:

Correct Option - 3


Detailed Solution:

S = 432 km/hr

tan 60° =   $\frac{h}{y}$

$h=\sqrt[]{3}$ y

$tan30°=\frac{h}{x+y}$           

$?x=432\times \frac{1000}{3600}\times 20$           

x = 2400 m

y = 1200 m

$\therefore h=1200\sqrt[]{3}m$          

 

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