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aeroplane flying horizontally 1km above the ground is observed at an elevation of 60 o .If after 10 seconds the elevation is observed to be 30 o .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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