Question

The angle of elevation of a jet plane from a point A on the ground is 60°. After 20 s at speed 432 km/h, the angle changes to 30°. If the jet plane is at constant height, then its height is :

• A. 3600√3 m
• B. 2400√3 m
• C. 1800√3 m
• D. 1200√3 m

Hint:

Always convert speeds to m/s ($1 \text{ km/h} = \frac{5}{18} \text{ m/s}$) when time is given in seconds and height is requested in meters.

Answer 1

The Correct Option is D

Step 1: Speed $= 432 \times \frac{5}{18} = 120 \text{ m/s}$. Distance $d = 120 \times 20 = 2400 \text{ m}$.
Step 2: Let height be $h$. In the first position, $\tan 60^\circ = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}}$.
Step 3: In the second position, $\tan 30^\circ = \frac{h}{x+2400} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{\frac{h}{\sqrt{3}} + 2400}$.
Step 4: $\frac{h}{\sqrt{3}} + 2400 = h\sqrt{3} \Rightarrow 2400 = h(\sqrt{3} - \frac{1}{\sqrt{3}}) = h(\frac{2}{\sqrt{3}})$.
Step 5: $h = \frac{2400\sqrt{3}}{2} = 1200\sqrt{3} \text{ m}$.