$=\sqrt{2 g h} \text {. }$

Range of water jet,

$x=\sqrt{2 g h} \times \text { Time of fall }$

$=\sqrt{2 g h} \times \frac{2 H}{\frac{2 H}{g}}=2 \sqrt{h H}$

Velocity $v$ of bucket is

$v=\frac{d x}{d t}=\frac{d}{d t} \cdot 2 \sqrt{h H}$

$=2 \sqrt{H} \cdot \frac{d \sqrt{h}}{d t}$

$\Rightarrow \quad v =2 \sqrt{H} \cdot \frac{1}{2 \sqrt{h}} \cdot \frac{d h}{d t}$
$=\sqrt{\frac{H}{h}} \cdot \frac{d h}{ d t}$   ............. $(i)$

Now, using equation of continuity at area $A_{0}$ and area $A$, we have

$A v_{1}=A_{0} V$

$\text { where, } v_{1} =\sqrt{2 g h}$

$\text { and } =\frac{d h}{d t}$

$\therefore A \sqrt{2 g h} =A_{0}\left(\frac{d h}{d t}\right)$

Substituting for $\frac{d h}{d t}$ from Eq $(i)$, we get

$v=\sqrt{\frac{H}{h}} \cdot \frac{A}{A_{0}} \cdot \sqrt{2 g h}$

$=\sqrt{\frac{5}{5}} \cdot\left(\frac{5}{500}\right) \cdot \sqrt{2 \times 10 \times 5}$

$=\frac{1}{10} \,ms ^{-1}$

$=0.1 \,ms ^{-1}$