Question

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°.

1. At what distance from the foot of the tree was he observing the bird sitting on the tree?
2. 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?
3. What is the speed of the bird in m/min if it had flown `20(sqrt3 + 1) m`?

Answer 1

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