91影视

The given data can be listed below,

• The height of the building is, $h=28.0\mathrm{m}$.
• The flow rate at which water enters the hose, $Av=0.500{\mathrm{m}}^3/\mathrm{s}$.

The mass of a flowing fluid will not change as it flows. A meaningful relationship that explains this is called the continuity equation. The continuity equation helps to explain the shape of a stream of fluid/liquid from a height.

By conservation of energy, initial speed and height water reaches can be expressed as,

$\frac{1}{2}m{v}^2=mgh$

Here, m is the mass of the water, v is the speed of flow, h is the height of the building and g is the acceleration due to gravity.

The speed of flow of water can be expressed by,

$$\begin{gathered} \left(\frac{v^2}{2}\right)=gh \\ {v}^2=2gh \\ v=\sqrt{2gh} \end{gathered}$$

Substitute values in the above equation.

$$\begin{gathered} v=\sqrt{2\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(28\mathrm{m}\right)} \\ =23.4\mathrm{m}/\mathrm{s} \end{gathered}$$

The volume flow rate of water can be expressed as,

$Av=\frac{蟺D^2}{4}v$

Here, A is the cross-section area of the nozzle and D is the diameter of the nozzle.

The diameter of the nozzle can be calculated as,

role="math" localid="1656495076166" $$\begin{gathered} \frac{蟺D^2}{4}v=0.500{\mathrm{m}}^3/\mathrm{s} \\ {\mathrm{D}}^2=\frac{4\left(0.500{\mathrm{m}}^3/\mathrm{s}\right)}{\mathrm{蟺惫}} \\ \mathrm{D}=\sqrt{\frac{4\left(0.500{\mathrm{m}}^3/\mathrm{s}\right)}{蟺v}} \end{gathered}$$

Substitute values in the above equation.

$$\begin{gathered} D=\sqrt{\frac{4\left(0.500{\mathrm{m}}^3/\mathrm{s}\right)}{蟺v}} \\ =\sqrt{\frac{4\left(0.500{\mathrm{m}}^3/\mathrm{s}\right)}{3.14\left(23.4\mathrm{m}/\mathrm{s}\right)}} \\ =0.165\mathrm{m} \end{gathered}$$

Thus, the maximum diameter of the nozzle is $0.165\mathrm{m}$.

The volume flow rate is given by Av. It is also expressed as $D^{2}v$which is constant.

So if D is twice as greater, then v is decreased by a factor of 4.

By conservation law of energy,

$h鈭�v^2$

By the above relation, h is decreased by a factor of 16. So, new h is calculated as,

$$\begin{gathered} h_1=\frac{28.0\mathrm{m}}{16} \\ =1.75\mathrm{m} \end{gathered}$$

Thus, the height water can reach when the diameter is twice its original value is $1.75\mathrm{m}$.