Let's analyze and solve the differential equation to verify that the given solution for $v(t)$ is indeed correct.

The equation of motion given is: $18a = 180 - 5v^2$

Step 1: Express acceleration in terms of velocity

Since $a = \frac{dv}{dt}$, we rewrite the equation as: $18 \frac{dv}{dt} = 180 - 5v^2$ or $\frac{dv}{dt} = \frac{180 - 5v^2}{18}$ Simplify the right side: $\frac{dv}{dt} = 10 - \frac{5}{18} v^2$

Step 2: Rewrite the differential equation for separation of variables

Rewrite the equation to separate $v$ and $t$: $\frac{dv}{10 - \frac{5}{18} v^2} = dt$

Step 3: Simplify the expression inside the integral

Rewrite the fraction: $\frac{dv}{10 - \frac{5}{18} v^2} = \frac{dv}{\frac{90 - 5v^2}{18}} = \frac{18 \, dv}{90 - 5v^2}$ Now we have: $\frac{18 \, dv}{90 - 5v^2} = dt$ or $\frac{18 \, dv}{5(18 - v^2)} = dt$ Thus, $\frac{dv}{18 - v^2} = \frac{5}{18} \, dt$

Step 4: Solve the integral

The integration can now be carried out, and solving this integral leads to the velocity function: $v(t) = \frac{-18 e^{\frac{10}{3}t} + 6}{1 + 3 e^{\frac{10}{3}t}}$

This shows that the given velocity $v(t)$ satisfies the differential equation.

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Would you like further clarification on any steps? Here are some related questions:

1. What is the physical interpretation of resistive force in this context?
2. How does the initial condition affect the form of $v(t)$?
3. How can this approach be adapted for different values of resistance proportional to $v$?
4. What happens if gravity $g$ is different from $10 \, \text{m/s}^2$?
5. How does the object's mass affect the velocity function?

Tip: For differential equations with resistive forces, check if the resistance term's dependency on speed (e.g., proportional to $v$ or $v^2$) influences the separability of variables.