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A bird is sitting on the top of a vertic...

A bird is sitting on the top of a vertical pole 20 m high and its elevation fron a point O on the ground is `45^@`,It Files off horizontally straight away from the point O.After one second , the elevation of the bird fron O is reduced to `30^@` Then the speed (in m/s) of the bird is

A

`40(sqrt(2)-1)`

B

`40(sqrt(3)-sqrt(2))`

C

`20sqrt(2)`

D

`20(sqrt(3)-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
In `Delta OA _1 B _ 1 `
` tan 45^(@) = (A_(1) B_(1))/(OB_(1)) rArr (20)/( OB_(1))=1 rArr =OB_(1)=20`

In ` Delta OA_(2) B_(2), tan 30^(@)=(20)/(OB_(2)) rArr OB_(2)=20sqrt(3)`
`rArr B_(1) B_(2) +OB_(1)=20 sqrt(3)`
`rArr B_(1)B_(2)=20sqrt(3)-20`
`rArr B_(1)B_(2)=20(sqrt(3)-1) m`
Now, Speed ` = (" Distance")/("Time") = 20(sqrt(3)-1) m//s`
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