A spherical tank has a circular orifice in its bottom through which the liquid flows out. The rate of change of the height $H$ as the liquid flows out through the hole can be estimated as:

$\frac{dH}{dt}=-\frac{CA\sqrt[\phantom{2}]{2gH}}{2rH-H^2}$

$\frac{2}{6}$

KIL$1005$

where $C=$ an empirically$-$derived coefficient, $A=$ the area of the orifice $(m^2),g=$ the gravitational constant $)=(9\mathrm{.}81\frac{m}{s^2},$ and $H=$ the height of liquid in the tank.

Use Midpoint method $($referring to the equation below$)$ to determine how long it will take for the water to flow out of a $4m$ diameter tank with an initial height of $0\mathrm{.}25m.$ Note that the orifice has a diameter of $3\mathrm{.}5cm$ and $C=0\mathrm{.}55.$ Use step size of $1$ minute $(60$ seconds$).$ Show the complete calculation steps. Plot the graph.

$y_{i+\frac{1}{2}}=y_i+f(x_i,y_i)\frac{h}{2}$

$y_{i+\frac{1}{2}}^{\text{'}}=f(x_{i+\frac{1}{2}},y_{i+\frac{1}{2}})$

$y_{i+1}=y_i+f(x_{i+\frac{1}{2}},y_{i+\frac{1}{2}})h$