A bird is sitting on the top of a vertical pole 20 m
high and its elevation from a point O on the ground is `45o`
. It flies off horizontally straight away from the
point O. After one second, the elevation of the bird from O is reduced to `30o`
. Then the speed (in m/s) of the bird is
(1) `40(sqrt(2)-1)`
(2)
`40(sqrt(3)-2)`
(3) `20""sqrt(2)`
(4) `20(sqrt(3)-1)`
A
`40 (sqrt(2)-1)`
B
`40 sqrt((3)-sqrt(2))`
C
`20 sqrt(2)`
D
`20(sqrt(3)-1)`
Text Solution
Verified by Experts
(4) `tan 30 ^2 =(20)/(20 + x )= 1/sqrt(3)` `rArr 20 + x = 20 sqrt( 3)` `rArr x= 20 ( sqrt( 3 -1) ` `Speed is 20 (sqrt(3-1)m//sec)`.
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