Question

38. The angle of elevation of a jet fighter plane from a point lying on the ground is 60°. After a flight of 15 seconds, the
angle of elevation changes to 30°. If the jet is flying at a speed of 720 km/h, find the constant height at which it is flying.
(Use V3 = 1.732)​

Answer 1

QUESTION:-

The angle of elevation of a jet fighter plane from a point lying on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet is flying at a speed of 720 km/h, find the constant height at which it is flying.

(Use √3 = 1.732)

SOLUTION:-

⇝Let A be the point of observation on the ground and B and C be the two positions of the jet. Let BL=CM=H metres, Let AL = x metres.

⇝speed of jet = 700km/hr = 700(5/18)

⇝.°. speed of jet = 200m/s

⇝ Time taken to cover the distance BC is 15sec.

⇝Distance BL=CM=200×15 = 3000m.

⇝ in ΔALB,

⇝ Cot60° = AL/BL

⇝1/√3 = x /h

⇝x = √3/h -------(1)

⇝ in ΔAMC,

⇝Cot30° = AM/MC

⇝ √3 = ( AL+LM/MC)

⇝x +3000 = h√3

⇝ x = h√3 - 3000 ---------(2)

⇝from equation, 1 & 2,

⇝√3h - 3000 = h/√3

⇝ 3h-3000 √3 = h

⇝ 3h -3000 ×1.732 =h (√3=1.732)

⇝2h = 5196

⇝h = 2598m (Answer)