Physics - Fluid Question with Solution | TestHub

As hole is made in the tank below the free surface of water, so water rushing from this hole follows a parabolic path.

The velocity of efflux liquid,

$\mathrm{v}=\sqrt{2\mathrm{gh}}$

Time,$\mathrm{t}=\sqrt{\frac{2\left(\mathrm{H}-\mathrm{h}\right)}{\mathrm{g}}}$

Horizontal range,$\mathrm{R}=\mathrm{vt}$

$\mathrm{ }\mathrm{R}={\left(2\mathrm{gh}\times \frac{2\left(\mathrm{H}-\mathrm{h}\right)}{\mathrm{g}}\right)}^{\frac{1}{2}}$

$\mathrm{ie}, {\mathrm{R}}^2=4\mathrm{h}\left(\mathrm{H}-\mathrm{h}\right)=4\left(\mathrm{Hh}-{\mathrm{h}}^2\right)$

The range is maximum if

$\frac{\mathrm{dR}}{\mathrm{dh}}=0$

$\mathrm{or} \frac{2\mathrm{RdR}}{\mathrm{dh}}=4\left(\mathrm{H}-2\mathrm{h}\right)$

$\mathrm{or} 0=\left(\mathrm{H}-2\mathrm{h}\right)$

$\mathrm{or} \mathrm{h}=\frac{\mathrm{H}}{2}$