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 45°. It flies off horizontally straight away from the point O. After one second, the elevation of the bird from O is reduced to 30°. Then the speed (in m/s) of the bird is
1
\(40(\sqrt{2}-1)\)
2
\(40(\sqrt{3}-\sqrt{2})\)
3
\(20 \sqrt{2}\)
4
\(20(\sqrt{3}-1)\)