Question

The angle of elevation of a jet fighter from a point A on the ground is 60°.
After flying 10 sec, the angle changes to 30° . If the jet is flying at a speed of
648 km/h, find the constant height at which the jet is flying.

Answer 1

Answer

The height at which the jet is flying is 900√3 metres.

Explanation

Let the distance between point A and the point B, lying vertically to the plane on the ground be x. (Refer the attachment)

The plane is flying at a speed of 648 km/hr = 648 × 5/18 m/s = 180 m/s

So, distance travelled by plane in 10 seconds = 180 × 10 = 1800 metres.

Let the height at which the jet is flying be h.

Now, in ∆ABY (refer the attachment)

tan(60°) = YB/AB

→ √3 = h/x

→ x = h/√3

In ∆ACM

tan(30°) = MC/AC

→ 1/√3 = h/(x + 1800)

→ x + 1800 = h√3

We know that x = h/√3

→ h/√3 + 1800 = h√3

Multiply both sides by √3

→ h + 1800√3 = 3h

→ 3h - h = 1800√3

→ 2h = 1800√3

→ h = 900√3 metres.