Explanation

(d) Let $\mathrm{\Delta}t$ be the time interval. Then,
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$2h = \left( u_{x} \right)(\mathrm{\Delta}t)$
or $u_{x} = \frac{2h}{\mathrm{\Delta}t}$ …(i)
Further, $h = u_{y}t - \frac{1}{2}gt^{2}$
or $gt^{2} - 2u_{y}t + 2h = 0$
$\therefore\ t_{1} = \frac{2u_{y} + \sqrt{4u_{y}^{2} - 8gh}}{2g}$
and $t_{2} = \frac{2u_{y} - \sqrt{4u_{y}^{2} - 8gh}}{2g}$
$\mathrm{\Delta}t = t_{1} - t_{2} = \frac{\sqrt{4u_{y}^{2} - 8gh}}{g}$
or $u_{y}^{2} = \frac{g^{2}(\mathrm{\Delta}t)^{2}}{4} + 2gh$
Given, $u_{x}^{2} + u_{y}^{2} = \left( 2\sqrt{gh} \right)^{2}$
$\therefore\ \frac{4h^{2}}{(\mathrm{\Delta}t)^{2}} + \frac{g^{2}(\mathrm{\Delta}t)^{2}}{4} + 2gh = 4gh$
$\frac{g^{2}}{4}(\mathrm{\Delta}t)^{4} - 2gh(\mathrm{\Delta}t)^{2} + 4h^{2} = 0$
$(\mathrm{\Delta}t)^{2} = \frac{2gh \pm \sqrt{4g^{2}h^{2} - 4g^{2}h^{2}}}{g^{2}/2} = \frac{4h}{g}$
or $\mathrm{\Delta}t = 2\sqrt{\frac{h}{g}}$