91Ó°ÊÓ

The diameter of pipe is \({d_{\rm{p}}} = 1.9\;{\rm{cm}}\).

The angle is \(\theta = 35^\circ \).

The water covers a radius of \(R = 6\;{\rm{m}}\).

The output diameter of each head is \({d_0} = 3\;{\rm{mm}}\).

In this problem, consider that the water is launched at ground level and used the relation of level range formula to find velocity of the water.

(a)

The relation of range level formula is given by,

\(\begin{array}{c}R = \frac{{{v^2}\sin 2\theta }}{g}\\v = \sqrt {\frac{{Rg}}{{\sin 2\theta }}} \end{array}\)

Here, \(g\) is the gravitational acceleration.

On plugging the values in the above relation,

\(\begin{array}{l}v = \sqrt {\left( {\frac{{\left( {6\;{\rm{m}}} \right)\left( {9.80\;{\rm{m/}}{{\rm{s}}^2}} \right)}}{{\sin 2\left( {35^\circ } \right)}}} \right)} \\v = 7.9\;{\rm{m/s}}\end{array}\)

Thus, the velocity of the water is \(7.9\;{\rm{m/s}}\).

(b)

The relation to find volume of water is given by,

\(\begin{array}{l}V = Av\\V = \left( {4\pi {r^2}} \right)v\\V = \left( {4\pi \frac{{{d_0}^2}}{4}} \right)v\\V = \left( {\pi {d_0}^2} \right)v\end{array}\)

Here, A is the area and V is the volume.

On plugging the values in the above relation.

\(\begin{array}{l}V = \left( {\pi \times {{\left( {3\;{\rm{mm}} \times \frac{{{{10}^{ - 3}}\;{\rm{m}}}}{{1\;{\rm{mm}}}}} \right)}^2}} \right)\left( {7.9\;{\rm{m/s}}} \right)\\V = \left( {2.22 \times {{10}^{ - 4}}\;{{\rm{m}}^{\rm{3}}}{\rm{/s}} \times \frac{{1\;{\rm{L/s}}}}{{1 \times {{10}^{ - 3}}\;{{\rm{m}}^{\rm{3}}}{\rm{/s}}}}} \right)\\V = 0.22\;{\rm{L/s}}\end{array}\)

Thus, the volume of water is \(0.22\;{\rm{L/s}}\).

(c)

The relation to find speed of water is given by,

\(\begin{array}{c}{\left( {Av} \right)_{\rm{s}}} = {\left( {Av} \right)_{\rm{h}}}\\{v_{\rm{s}}} = \frac{{{{\left( {Av} \right)}_{\rm{h}}}}}{{{A_{\rm{s}}}}}\\{v_{\rm{s}}} = \frac{{{{\left( {Av} \right)}_{\rm{h}}}}}{{\frac{{\pi {d_{\rm{p}}}^2}}{4}}}\end{array}\)

On plugging the values in the above relation.

\(\begin{array}{l}{v_{\rm{s}}} = \frac{{2.22 \times {{10}^{ - 4}}\;{{\rm{m}}^{\rm{3}}}{\rm{/s}}}}{{\left( {\frac{{\pi {{\left( {1.9\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)}^2}}}{4}} \right)}}\\{v_{\rm{s}}} = 0.78\;{\rm{m/s}}\end{array}\)

Thus, the speed of water flowing is \(0.78\;{\rm{m/s}}\).