A stone is projected with velocity 2 gh , so that it just clears...

2h = ut$</p><p>${u}_{x} = \dfrac{2h}{\Deltat}\left[∵ u = {u}_{x}, t = \Deltat\right] . . . . \left(i\right)$</p><p>$h = {u}_{y}t - \dfrac{1}{2}g{t}^{2}$</p><p>$g{t}^{2} - 2{u}_{y}t + 2h = 0$</p><p>${t}_{1} = \dfrac{2{u}_{y} + \sqrt{4{u}_{y}^{2} - 8gh}}{2g}$</p><p>${t}_{2} = \dfrac{2{u}_{y} - \sqrt{4{u}_{y}^{2} - 8gh}}{2g}$</p><p>$\Deltat = {t}_{1} - {t}_{2} = \dfrac{\sqrt{4{u}_{y}^{2} - 8gh}}{g}$</p><p>${u}_{y}^{2} = \dfrac{{g}^{2} ( \Deltat{)}^{2}}{4} + 2gh$</p><p>${u}_{x}^{2} + {u}_{y}^{2} = {u}^{2} = ( 2\sqrt{gh}{)}^{2}$</p><p>$\dfrac{4{n}^{2}}{( \Deltat{)}^{2}} + \dfrac{{g}^{2} ( \Deltat{)}^{2}}{4} + 2gh = 4gh$</p><p>$\dfrac{{g}^{2}}{4}{\left(\Deltat\right)}^{4} - 2gh{\left(\Deltat\right)}^{2} + 4{h}^{2} = 0$</p><p>\left(\Deltat\right) = \dfrac{2gh \pm \sqrt{4{g}^{2}{h}^{2} - 4{g}^{2}{h}^{2}}}{{g}^{2} / 2} = \dfrac{4h}{g}$

$\Deltat = 2\sqrt{\dfrac{h}{g}}$