Q. An aircraft is flying at a height of $34000m$ above the ground. If the angle subtended at a ground observation point by the aircraft positions $10s$ apart is $30{^\circ}$, then the speed of the aircraft is
A) $19.63ms^{- 1}$
B) $1963\ ms^{- 1}$
C) $108\ ms^{- 1}$
D) $196.3\ ms^{- 1}$
Explanation
(d) [image:3_image203.emf]
$O$ is the observation point at the ground. $A$ and $B$ are the positions of aircraft for which $\angle AOB = 30{^\circ}$. Time taken by aircraft from $A$ to $B$ is $10s$
$\mathrm{\Delta}AOB$
$\tan{30{^\circ}} = \frac{AB}{3400}$
$AB = 3400\tan{30{^\circ}} = \frac{3400}{\sqrt{3}}m$
$\therefore$ Speed of aircraft,
$v = \frac{AB}{10} = \frac{3400}{10\sqrt{3}} = 196.3\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.