Question

An aircraft is flying at a height of 3400 m above the ground. It the angle subtended at a ground observation point by the aircraft positions 10 s apart is $30 ^ { \circ }$ , then the speed of the aircraft is

• A. $10.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$(10.8 m/s)
• B. $1963 \mathrm {~m} \mathrm {~s} ^ { - 1 }$(1963 m/s)
• C. $108 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (108 m/s)
• D. $196.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (196.3 m/s)

Answer 1

Answer: (D)

$196.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (196.3 m/s)

Answer 2

O is the observations points at the grounds, A and B are the positions of aircraft for which $\angle \mathrm { AOB } = 30 ^ { \circ }$, Time taken by aircraft from A to B is 10s.

In

$\tan 30 ^ { \circ } = \frac { \mathrm { AB } } { 3400 }$

$\mathrm { AB } = 3400 \tan 30 ^ { \circ } = \frac { 3400 } { \sqrt { 3 } } \mathrm {~m}$

$\therefore$Speed of aircraft.