An aeroplane is flying horizontally 1 km above the ground is observed at an elevation of 60. If after 10 seconds the elevation is observed to be 30, the uniform speed of the aeroplane is_______

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Solution

240 $\sqrt{3}$ km/h
v = s/t = $\frac{1. c o t 30^{0} - 1. c o t 60^{0}}{10 / 3600}$
= $\frac{2}{\sqrt{3}}$ (360) = 240 $\sqrt{3}$ km/h

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An aeroplane is flying horizontally 1 km above the ground is observed at an elevation of 60. If after 10 seconds the elevation is observed to be 30, the uniform speed of the aeroplane is_______

240 √3 km/hv = s/t = 1.cot300−1.cot60010/3600= 2√3 (360) = 240√3 km/h

240 $\sqrt{3}$ km/h
v = s/t = $\frac{1. c o t 30^{0} - 1. c o t 60^{0}}{10 / 3600}$
= $\frac{2}{\sqrt{3}}$ (360) = 240 $\sqrt{3}$ km/h