An observer spots a plane travelling at an altitude of 4000m with an angle of elevation of 30 degrees at 11:59 am. At noon, the plane is directly over him. What is the speed of the plane m/s?
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At 11:59 , let the horizontal distance of the plane from the observer be x m
then tan30° = 4000/x
x = 4000/tan30° = 6928.203... m
so in 1 min (11:59 - 12:00) it travelled 6928.203... m
1 min = 60 s
speed = distance/time = 6928.203../60 m/s
= appr 115.5 m/s