When a stone is dropped from a height h, it hits the ground with a momentum

$P = m\sqrt {2gh}$     ....(1)

where m is the mass of the stone.
When the same stone is dropped from a height 2h (i.e. 100% of initial), then its momentum with which it hits the ground becomes

$P' = m\sqrt {2g\left( {2h} \right)} = \sqrt 2 P$     [Using (1)]     ...(2)

% change in momentum = ${{P' - P} \over P} \times 100\%$

$= {{\sqrt 2 P - P} \over P} \times 100\% = 41\%$