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aeroplane flying horizontally 1km above the ground is observed at an elevation of 60o .If after 10 seconds the elevation is observed to be 30o .The uniform speed for an hour of the aeroplane is
When the elevation was 60, distance of the point of observation to the point on the groundbelow the plane = 1xcot60 = 1/sqrt(3) km Similarly when the angle becomes 30, distance becomes = 1xcot30 = sqrt(3) km So, distance travelled = sqrt 3 -1/sqrt 3 km = 2/sqrt3 km time taken = 10sec = 10/3600 hr hence speed = distance / time = 2x3600/10x sqrt(3) = 720/sqrt (3) km/hr = 415.7 km/hr
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