(c) Vertical distance covered by water before striking ground$= (H - h)$. Time taken is, $t = \sqrt{2(H - h)g}$; Horizontal velocity of water coming out of hole at $P,\ u = \sqrt{2gh}$
$\therefore$ Horizontal range
$= ut = \sqrt{2gh} = \sqrt{2(H - g)/g}$
$= 2\sqrt{h(H - h)}$