Question

An aircraft is flying at a uniform speed of $200m{s}^{-1}$. If the angle subtended at an observation point on the ground by two positions of the aircraft $10s$ apart is ${30}^0$, the height of the aircraft above the ground is:( $\tan {15}^0$ = 0.26795 )

Note that the observer is below the midpoint of the path covered by the aircraft in $10s$.

• A. $3.73Km$
• B. $7.73Km$
• C. $2.73Km$
• D. $1.73Km$

Answer 1

The correct option is A $3.73Km$

Aircraft is in uniform motion at $200m{s}^{-2}$ for $10s$ $⟹$ it travelled $200\times 10=2000m$.

Rest of the problem is purely geometrical.
Let the aircraft flew from $A$ to $B$. Let observer be sitting at $O$ directly below point $P$ which is is the midpoint of $AB$.
Distance travelled $AB=2km$

Hence $AP=PB=1km$
AB subtends ${30}^{\circ }$ at $O$ $⟹\angle AOB={30}^{\circ }$
Since $△OPB$ is a right angled triangle with $\angle BOP={15}^{\circ }$, $\frac{PB}{OP}=\tan {15}^{\circ }$
Hence, the height of aircraft above the ground $OP=\frac{PB}{\tan {15}^{\circ }}=\frac{1}{\tan {15}^{\circ }}km=3.73km$