Step 1

We are given the angle of the water ejected from the fountain is $\theta = 40^\circ $.

We are asked to determine the maximum and minimum speed at which water can be ejected from the nozzle so that it does not splash over the sides of the basin at B and C.

 
Step 2

The diagram of the system is shown as:

We have the distance between nozzle and point B is ${x_1} = 100\;{\rm{mm}} \times \left( {\frac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}}} \right) = 0.1\;{\rm{m}}$.

We have the distance between nozzle and point C is ${x_2} = 100\;{\rm{mm}} + 250\;{\rm{mm}} = 350\;{\rm{mm}}\left( {\frac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}}} \right) = 0.35\;{\rm{m}}$.

We have the vertical displacement of water is $y = 50\;{\rm{mm}}$.

 
Step 3

The formula to calculate the range of water ejected by nozzle is,

\[R = \left( {{v_A}\sin \theta } \right)t\]
 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} R = \left( {{v_A}\sin 40^\circ } \right)t\\ R = 0.643{v_A}t\\ t = \frac{R}{{0.64{v_A}}} \end{array}\]
 
Step 5

The formula to calculate the vertical displacement of the ejected water is,

\[y = \left( {{v_A}\cos \theta } \right)t + \frac{1}{2}g{t^2}\]

Here, g is the gravitational acceleration, and its standard value is $9.81\;{\rm{m/}}{{\rm{s}}^2}$

 
Step 6

Substitute the value of t in the above expression.

\[\begin{array}{c} - 50\;{\rm{mm}} \times \left( {\frac{{1\;{\rm{m}}}}{{1000\;{\rm{mm}}}}} \right) = \left( {{v_A}\cos 40^\circ } \right)\left( {\frac{R}{{0.64{v_A}}}} \right) + \frac{1}{2}\left( { - 9.81\;{\rm{m/}}{{\rm{s}}^2}} \right){\left( {\frac{R}{{0.64{v_A}}}} \right)^2}\\ {\left( {{v_A}} \right)^2} = \frac{{4.905{R^2}}}{{0.64\left[ {0.76R + 0.05\left( {0.64} \right)} \right]}}\\ {v_A} = \sqrt {\frac{{4.905{R^2}}}{{0.48R + 0.0205}}} \end{array}\]
 
Step 7

Substitute the value $R = {x_1} = 0.1\;{\rm{m}}$ in above expression to obtain minimum velocity.

\[\begin{array}{l} {\left( {{v_A}} \right)_{\min }} = \sqrt {\frac{{4.905{{\left( {0.1} \right)}^2}}}{{0.48\left( {0.1} \right) + 0.0205}}} \\ {\left( {{v_A}} \right)_{\min }} = \sqrt {0.716} \\ {\left( {{v_A}} \right)_{\min }} = 0.84\;{\rm{m/s}} \end{array}\]
 
Step 8

Substitute the value $R = {x_2} = 0.35\;{\rm{m}}$ in above expression to obtain maximum velocity.

\[\begin{array}{l} {\left( {{v_A}} \right)_{\max }} = \sqrt {\frac{{4.905{{\left( {0.35} \right)}^2}}}{{0.48\left( {0.35} \right) + 0.0205}}} \\ {\left( {{v_A}} \right)_{\max }} = \sqrt {3.1876} \\ {\left( {{v_A}} \right)_{\max }} = 1.78\;{\rm{m/s}} \end{array}\]