Question

the angle of elevation of a jet plane from a point A on the ground is 60°. After the flight of 30 seconds the angle of elevation changes to 30°. If the jet plane is flying at a constant height of 3600√3m. Find the speed of the jet plane.​

Answer 1

In triangle ACE,

\tan(30) = \frac{ac}{ce} \\ \frac{1}{ \sqrt{3} } = \frac{3600 \sqrt{3} }{ce} \\ ce = 3600 \sqrt{3} \times \sqrt{3} \\ ce = 10800 \: m

CE = 10800 m

AC = BD =

3600 \sqrt{3} \: m

In triangle BED,

\tan(60) = \frac{bd}{de} \\ \sqrt{3} = \frac{3600 \sqrt{3} }{de} \\ de = \frac{3600 \sqrt{3} }{ \sqrt{3} } = 3600 \: m

CD + DE = CE

CD + 3600 = 10800

CD = 10800 - 3600 = 7200 m

Distance travelled = 7200 m

Time taken = 30 seconds

speed = \frac{distance}{time} = \frac{7200}{30} = 240 \: m {s}^{ - 1}

In 1 second = 240 m

In 3600 seconds (1 hour) = 240 × 3600 = 864000 m = 864 km

In hour = 864 km

Speed = 864 km/h

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SAKNA1

SAKNA1 Ace

In triangle ACE,

CE = 10800 m

AC = BD =

In triangle BED,

CD + DE = CE

CD + 3600 = 10800

CD = 10800 - 3600 = 7200 m

Distance travelled = 7200 m

Time taken = 30 seconds

In 1 second = 240 m

In 3600 seconds (1 hour) = 240 × 3600 = 864000 m = 864 km

In hour = 864 km

Speed = 864 km/h

Answer 2

In triangle ACE,

CE = 10800 m

AC = BD =

In triangle BED,

CD + DE = CE

CD + 3600 = 10800

CD = 10800 - 3600 = 7200 m

Distance travelled = 7200 m

Time taken = 30 seconds

In 1 second = 240 m

In 3600 seconds (1 hour) = 240 × 3600 = 864000 m = 864 km

In hour = 864 km

Speed = 864 km/h

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