Question

An aeroplane flying horizontally $1$, above the ground is observed at an elevation of $60$ and after $10$ seconds the elevation is observed to be $30$. The uniform speed of the aeroplane in $/h$ is-

• A. $240$
• B. $240\sqrt{3}$
• C. $60\sqrt{3}$
• D. None of these

Answer 1

The correct option is B $240\sqrt{3}$

At $t=0$ plane is at A and at $t=10$, the plane is at B. Distance covered$=AB$

In $△AHO$

$\tan 60°=\frac{OH}{AH}$

$AH=\frac{\left(1000\right)}{\sqrt{3}}m$

In $△BHO$

$\tan 30°=\frac{OH}{BH}$

$BH=\sqrt{3}\left(1000\right)m$

$\therefore AB=BH-AH=(\sqrt{3}-\frac{1}{\sqrt{3}})\left(1000\right)m$

$=\frac{2}{\sqrt{3}}.1000m$

Speed$=\frac{Distance}{Time}$

Speed$=\frac{2}{\sqrt{3}}\times \frac{1000}{10}㎧$

$=\frac{200}{\sqrt{3}}㎧$

$=\frac{200}{\sqrt{3}}\times \frac{18}{5}㎞h^{-1}$

$=40\times 6\sqrt{3}$

Speed$=240\sqrt{3}㎞h^{-1}$