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A bird is sitting on the top of a vertical pole 20 mhigh and its elevation from a point O on the ground is `45o`. It flies off horizontally straight away from thepoint 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)` |
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Answer» in `/_ AOB` `tan 45^@ = (AB)/(0B)` `1= 20/(OB)` `OB= 20m` in`/_ COD` `tan 30^@ = (CD)/(OD)` `1/sqrt3 = 20/(20+x)` `20 + x = 20 sqrt3` `x= 20(sqrt3 - 1)` Answer |
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