The angle of elevation of a jet fighter from a point A on the ground is x. After t seconds, the angle of elevation changes to y. If the velocity of jet is v km/hr, prove that constant height is 5vt.tanx.tany/18(tanx - tany)
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H = 1500√3 m = 1500×1.732 = 2598 m. ... The angle of elevation of a jet fighter from a point A on the ground is 60 degree . after a flight of 15 seconds the angle of elevation changes to 30 degree .if the jet is flying at the speed 720 km/h find the constant height at which the jet is flying.
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