Answer
$2.00$
Work Step by Step
We know that
$D=vt$
$\implies D=(\sqrt{2gh})(\sqrt{\frac{2H}{g}})$
$D=2\sqrt{hH}$......eq(1)
It is given that both $h$ and $H$ are increased by a factor of 2
$\implies D=2\sqrt{(2h)(2H)}$
$D=2\sqrt{4hH}$
$D=2(2\sqrt{hH})=2D$
Thus, the distance is increased by a factor of 2.
