and time taken by liquid to reach the ground

$t = \sqrt {\frac{{2(h + h - y)}}{g}} = \sqrt {\frac{{2(2h - y)}}{g}} $

$\therefore$ Horizontal distance covered by liquid

$x = v.t. = \sqrt {2gy} \times \sqrt {\frac{{2(2h - y)}}{g}} = \sqrt {4y(2h - y)} $

==> ${x^2} = 4y(2h - y)$

==> $\frac{{d{{(x)}^2}}}{{dy}} = 8h - 8y$

for $x$ to be maximum, $\frac{d}{{dy}}({x^2}) = 0$

$\therefore 8h - 8y = 0$ or $h = y$

So ${x_m} = \sqrt {4h(2h - h)} = 2h$