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

To solve this problem, we need to calculate the height of the jet plane using trigonometry. The jet is flying at a constant height, and the change in angles of elevation gives us the information needed to find this height.

The given data includes:

• The initial angle of elevation from point A is 60°.
• The angle of elevation changes to 30° after 20 seconds.
• The speed of the jet is 432 km/h, which is equivalent to 432 \times \frac{1000}{3600}\, m/s = 120 \, m/s.

Given that the jet covers a horizontal distance of D in 20 seconds:

D = 120 \, m/s \times 20 \, s = 2400 \, m

Using trigonometry, let h be the height of the jet, and AB and AC be the initial and final horizontal distances from point A (on the ground) to the vertical under the jet.

At angle 60^\circ:

\tan 60^\circ = \frac{h}{AB} , where \tan 60^\circ = \sqrt{3}. Hence,

h = \sqrt{3} \cdot AB \, \ldots (1)

At angle 30^\circ:

\tan 30^\circ = \frac{h}{(AB + 2400)} , where \tan 30^\circ = \frac{1}{\sqrt{3}}. Hence,

h = \frac{1}{\sqrt{3}} \cdot (AB + 2400) \, \ldots (2)

We equate equations (1) and (2) to solve for h:

\sqrt{3} \cdot AB = \frac{1}{\sqrt{3}} \cdot (AB + 2400)

Multiplying throughout by \sqrt{3}, we get:

3AB = AB + 2400

Rearranging the terms gives:

2AB = 2400

AB = 1200 \, m

Substituting AB into equation (1):

h = \sqrt{3} \cdot 1200 = 1200 \sqrt{3} \, m

Thus, the height of the jet plane is 1200√3 m.