Question

A body falling freely from rest has a velocity v after it has fallen through a distance h..the distance it has to fall down further for its velocity to become 2v is .....times h. 

Answer 1

use $v^2-u^2=2gs$
intial velocity u=0
at velocity=v height=h
therefore
$v^2=2gh$
then let intial velocity u=v
and final velocity=2v
therefore
$(2v)^2-v^2=2gs$
$4v^2-v^2=3v^2=6gh=2g(3h)$
the distance it has to fall down further for its velocity to become 2v is 3.times h.