Question

The angle of elevation of a jet plane from a point ${}^{\prime }{A}^{\prime }$ on the ground is ${60}^o$. After a flight of $15seconds$, the angle of elevation changes to ${30}^o$. If the jet plane is flying at a constant height of $1500\sqrt{3}meters$. Find the speed of jet plane $(\sqrt{3}=1.732)$.

• A. $200m/s$
• B. $600m/s$
• C. $300m/s$
• D. $400m/s$

Answer 1

Answer: (A)

$200m/s$

Let D and E be the initial and final positions of the plane respectively.

Consider the $△ABD$

$\tan 60=\frac{BD}{AB}$ ............ $\because tan\left(\Theta \right)=\frac{Opposite}{Adjacent}$

$\sqrt{3}=\frac{1500\sqrt{3}}{AB}$

$AB=1500$

Consider the $△ACE$

$\tan 30=\frac{CE}{AC}$

$\frac{1}{\sqrt{3}}=\frac{1500\sqrt{3}}{AC}$

$AC=4500$

$BC=AC-AB=4500-1500=3000m$

$DE=BC=3000m$

i.e. the jet plane travels a distance of $3000m$ in $15$ sec

Therefore,

Speed of the jet plane$=\frac{3000}{15}=200m/s$