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Question
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One evening, Kaushik was in a park. Children were playing cricket. Birds were singing on a nearby tree of height 80m. He observed a bird on the tree at an angle of elevation of 45°. When a sixer was hit, a ball flew through the tree frightening the bird to fly away. In 2 seconds, he observed the bird flying at the same height at an angle of elevation of 30° and the ball flying towards him at the same height at an angle of elevation of 60°.
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- At what distance from the foot of the tree was he observing the bird sitting on the tree?
- How far did the bird fly in the mentioned time?
(or)
After hitting the tree, how far did the ball travel in the sky when Kaushik saw the ball? - What is the speed of the bird in m/min if it had flown `20(sqrt3 + 1) m`?
Solution
i. tan 45° = `80/(CB)`
`\implies` CB = 80 m
ii. tan 30° = `80/(CE)`
`\implies 1/sqrt(3) = 80/(CE)`
`\implies` CE = `80sqrt(3)`
Distance the bird flew = AD = BE = CE – CB = `80sqrt(3) - 80 = 80(sqrt(3) - 1)m`
(or)
tan 60° = `80/(CG)`
`\implies sqrt(3) = 80/(CG)`
`\implies` CG = `80/sqrt(3)`
Distance the ball travelled after hitting the tree = FA = GB = CB – CG
GB = `80 - 80/sqrt(3)`
= `80(1 - 1/sqrt(3))m`
iii. Speed of the bird = `"Distance"/"Time taken"`
= `(20(sqrt(3) + 1))/2` m/sec
= `(20(sqrt(3) + 1))/2 xx 60` m/min
= `600(sqrt(3) + 1)` m/min
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