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Grade: 12

                        

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

2 years ago

Answers : (1)

Arun
24735 Points
							
When the elevation was 60, distance of the point of observation to the point on the ground
below 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
2 years ago
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