Question

The angle of elevation of a jet plane from a point A on the ground is ${60}^0$. After a flight of 15 seconds, the angle of elevation changes to ${30}^0$. If the plane at a constant height of $1500\sqrt{3}m$, then the speed of jet plane is :

• A. 200 m/s
• B. 1980 m/s
• C. 240 m/s
• D. 220 m/s

Answer 1

The correct option is A 200 m/s

$BE=CD=1500\sqrt{3}m,BC=ED$

Let the speed be ${}^{\prime }{V}^{\prime }$ m/s

Time taken$=15$s

Then distance $BC=15V$m$=ED$

In $△ABE$

$\tan {60}^{\circ }=\frac{BE}{AE}\Rightarrow \sqrt{3}=\frac{1500\sqrt{3}}{AE}\Rightarrow AE=1500$m

In $△ACD$

$\tan {30}^{\circ }=\frac{CD}{AD}\Rightarrow \frac{1}{\sqrt{3}}=\frac{1500\sqrt{3}}{AE+ED}\Rightarrow 1500\sqrt{3}\times \sqrt{3}=1500+ED\Rightarrow 3\times 1500-1500=EDED=4500-1500=3000m$

Now $15V=3000\Rightarrow V=\frac{3000}{15}m/s=200m/s$