Given $k = 0.15 \, \text{m/min}$ and the initial condition $h(0) = 0$, we can solve the integral to find the function $h(t)$.

Restating the Differential Equation:

From the previous part, we have:

$\frac{dh}{dt} = \frac{16}{\pi h^2} \left(3 - 0.15 h^3\right)$

Separating the variables:

$h^2 \, \frac{dh}{3 - 0.15 h^3} = \frac{16}{\pi} \, dt$

Solve the Integral:

To solve this, we integrate both sides. Let's start with the left side:

$\int \frac{h^2}{3 - 0.15 h^3} \, dh = \frac{16}{\pi} \int dt$

For simplicity, substitute $u = 3 - 0.15 h^3$. Then:

$du = -0.45 h^2 \, dh$

So, the integral becomes:

$-\frac{1}{0.45} \int \frac{1}{u} \, du = \frac{16}{\pi} t + C$

Integrating:

$-\frac{1}{0.45} \ln |u| = \frac{16}{\pi} t + C$

Substituting back $u = 3 - 0.15 h^3$:

$-\frac{1}{0.45} \ln |3 - 0.15 h^3| = \frac{16}{\pi} t + C$

Simplify this equation:

$\ln |3 - 0.15 h^3| = -0.45 \cdot \frac{16}{\pi} t - 0.45C$

$\ln |3 - 0.15 h^3| = -\frac{7.2}{\pi} t + C_1$

where $C_1$ is a constant.

Exponentiating Both Sides:

Exponentiate both sides to solve for $h^3$:

$|3 - 0.15 h^3| = e^{C_1} e^{-\frac{7.2}{\pi} t}$

Let $e^{C_1} = A$ (a new constant):

$3 - 0.15 h^3 = A e^{-\frac{7.2}{\pi} t}$

Solve for $h^3$:

$h^3 = \frac{3 - A e^{-\frac{7.2}{\pi} t}}{0.15}$

$h^3 = \frac{20}{3} - \frac{A}{0.15} e^{-\frac{7.2}{\pi} t}$

Taking the cube root:

$h(t) = \left(\frac{20}{3} - \frac{A}{0.15} e^{-\frac{7.2}{\pi} t}\right)^{1/3}$

Determine $A$ Using Initial Conditions:

Given $h(0) = 0$:

$0 = \left(\frac{20}{3} - \frac{A}{0.15}\right)^{1/3}$

Thus,

$\frac{20}{3} = \frac{A}{0.15}$

$A = 0.15 \times \frac{20}{3} = 1$

Finally, the function $h(t)$ becomes:

$h(t) = \left(\frac{20}{3} - \frac{1}{0.15} e^{-\frac{7.2}{\pi} t}\right)^{1/3}$

or simplified:

$h(t) = \left(\frac{20}{3} - \frac{20}{3} e^{-\frac{7.2}{\pi} t}\right)^{1/3}$

This is the height of water in the tank at any time $t$.

Would you like more details or have any questions?

Further questions:

1. What would happen if the proportionality constant $k$ were larger or smaller?
2. How does this function behave as $t$ becomes very large?
3. Could we find an approximate time at which the tank is nearly full?
4. How would this function change if the rate of water input varied over time?
5. What are the limitations of this model in real-world applications?

Tip: When working with differential equations, always check your solution by plugging it back into the original equation to ensure it satisfies all conditions.