Question

The angle of elevation of a jet plane from a point A on the ground is 60∘. After a flight of 15 second, the angle of elevation changes to 30∘. If the jet plane is flying at a constant height of  15002​m.Find the speed of the jet plane.

Answer 1

Correct Question :

The angle of elevation of a jet plane point A on the ground is 60°. After a flight of 15 second, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 1500√3m. Find the speed of the jet plane.

AnsWer :

200m/s

Solution :

Let PQ distance be x m.

Height 1500√3 m.

Now,

In ∆TQP,

$\tt \tan \theta = \frac{TQ}{PQ} \\ \tt \tan60 \degree = \frac{1500 \sqrt{3} }{x}$

$\tt \cancel{\sqrt{3}} \degree = \frac{1500 \cancel{\sqrt{ 3} }}{x} \\ \tt x = 1500 \: m.$

Again,In ∆PRS

$\tt \tan \theta = \frac{SR}{PR} \\ \tt \tan30 \degree = \frac{1500}{PR}$

$\tt \frac{1}{ \sqrt{3} } = \frac{1500 \sqrt{3} }{PR} \\ \tt PR = 4500.$

So, we can say that,

$\tt PR = PQ + QR. \\ \tt4500 = x + QR. \\ \tt QR = 4500 - 1500. \\ \tt QR = 3000 \: m.$

Now,

We know the formula,

Distance =Speed * Time.

Apply,

$\tt D = S \times T. \\ \tt3000 = 15s \\ \tt s = \frac{ \cancel{30}00}{ \cancel{15}} \\ \tt s = 200 \: m /s.$

Therefore,the speed of jet plane is 200 m/s.