Solveeit Logo

Question

Physics Question on Motion in a plane

An aircraft is flying at a height of 3400m3400\, m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10s10\,s apart is 3030^\circ then the speed of the aircraft is

A

19.63ms119.63\,m{{s}^{-1}}

B

196.3ms1196.3\,m{{s}^{-1}}

C

108ms1108\,m{{s}^{-1}}

D

1963ms11963\,m{{s}^{-1}}

Answer

196.3ms1196.3\,m{{s}^{-1}}

Explanation

Solution

OO is the observation point at the ground. AA and BB are the positions of aircraft for which ?AOB=30?? AOB=30^?.
Time taken by aircraft from AA to BB is 10s10\,s.
In ΔAOB,\Delta AOB,
tan30?tan\,30^{?}
=AB3400=\frac{AB}{3400}
AB=3400tan30?AB=3400\,tan\,30^{?}
=34003m=\frac{3400}{\sqrt{3}}m
\therefore Speed of aircraft, v=AB10v=\frac{AB}{10}
=3400103=\frac{3400}{10\sqrt{3}}
=196.3ms1=196.3\,ms^{-1}