As per the given problem there are two holes in the side walls at heights of $h1$ and $h2$ respectively such that the range of efflux at the bottom of the vessel is same. Hence,

$\left.\left.x=2 \sqrt{(} H-h_{1}\right) h_{1}=2 \sqrt{(} H-h_{2}\right) h_{2}$

$\Rightarrow\left(H-h_{1}\right) h_{1}=\left(H-h_{2}\right) h_{2}$

$\therefore H h_{1}-\left(h_{1}\right)^{2}=H h_{2}-\left(h_{2}\right)^{2}$

$\therefore H\left(h_{1}-h_{2}\right)=\left(h_{1}\right)^{2}-\left(h_{2}\right)^{2}$

$\therefore H=\frac{\left(h_{1}\right)^{2}-\left(h_{2}\right)^{2}}{h_{1}-h_{2}}$

$\therefore H=\frac{\left(h_{1}\right)^{2}-\left(h_{2}\right)^{2}}{h_{1}-h_{2}} \frac{\left(h_{1}+h_{2}\right)}{\left(h_{1}+h_{2}\right)}$

$\Rightarrow H=h_{1}+h_{2}$

For maximum range the height will be$:$ $H=\frac{h}{2}$

Hence, $H=\frac{h_{1}+h_{2}}{2}$