-y^2 =25y+6y^3find y
2 answers:
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Answer:
The horizontal distance from the plane to the person on the runway is 20408.16 ft.
Step-by-step explanation:
Consider the figure below,
Where AB represent altitude of the plane is 4000 ft above the ground , C represents the runner. The angle of elevation from the runway to the plane is 11.1°
BC is the horizontal distance from the plane to the person on the runway.
We have to find distance BC,
Using trigonometric ratio,
Here, ,Perpendicular AB = 4000
Solving for BC, we get,
(approx)
(approx)
Thus, the horizontal distance from the plane to the person on the runway is 20408.16 ft
