Follow the link http://img684.imageshack.us/img684/8819/unledaj.png

angle AOB=45, angle COD=30. AB=CD, CA=DB. A is the initial position of the plane; and C is the position of the plane after 20 seconds. In 20 seconds, the plane covers a distance of 2000 mitres. So AC=2000.

tan(AOB)=tan(45)=1=AB/BO. So AB=BO

tan(COD)=tan(30)=1/sqrt(3)=CD/DO=CD/(DB+BO)=CD/(2000+BO)=AB/(2000+BO)=AB/(2000+AB)

That is AB/(2000+AB) = 1/sqrt(3)

Solving, AB = 2000/{sqrt(3)-1} which is the required height.

EDIT: In case the above link is non-accessible, here is another link http://oi56.tinypic.com/2eldz4n.jpg