Given is a balloon that is filled with hot air and is fixed until t=0 at the ground. $\theta(t)$ defines the temperature of the air inside the balloon and $v(t)$ defines the climb rate. The following model describes the balloon:
$\dot{\theta}(t) = -0.1\cdot (\theta(t) - \theta_U) $
$\dot{v}(t) = -0.5\cdot v(t) + 2 \cdot (\theta(t) - \theta_U) $
$\theta_U$ is the constant ambient temperature.
Question: The maximum height that the ballon can reach depends on the beginning temperature of the air in the ballon $\theta_0 = \theta(t=0)$ and $\theta_U$. The balloon starts once with $\theta_U = 10$ and once with $\theta_U = 0$. If both times he reaches the same maximum height, what is the difference between the starting temperatures $\theta_0$ each time?
My ideas: I already managed to convert the equations into state space and to draw a simulink graph. I tried to transform the equations into complex variable domain and to integrate $\dot{v}(t)$. None of that really worked. I guess I must find out a formulation for the maximum height depending on $\theta_U$ and $\theta_0$. What I know for sure is that the ballon will stop rising when $\theta_U = \theta(t)$, and on that time is $v(t) = 0$.
Do you have any indeas how to find out the maximum height?
Well, we can take the Laplace transform of the system of equations:
$$ \begin{cases} \theta'\left(t\right)=-\frac{1}{10}\cdot\left(\theta\left(t\right)-\theta_\text{u}\right)\\ \\ \text{v}'\left(t\right)=-\frac{1}{2}\cdot\text{v}\left(t\right)+2\cdot\left(\theta\left(t\right)-\theta_\text{u}\right) \end{cases}\tag1 $$
So, we get:
$$ \begin{cases} \text{s}\cdot\Theta\left(\text{s}\right)-\theta\left(0\right)=-\frac{1}{10}\cdot\left(\Theta\left(\text{s}\right)-\frac{1}{\text{s}}\cdot\theta_\text{u}\right)\\ \\ \text{s}\cdot\text{V}\left(\text{s}\right)-\text{v}\left(0\right)=-\frac{1}{2}\cdot\text{V}\left(\text{s}\right)+2\cdot\left(\Theta\left(\text{s}\right)-\frac{1}{\text{s}}\cdot\theta_\text{u}\right) \end{cases}\tag2 $$
We can also set:
$$ \begin{cases} \Theta\left(\text{s}\right)=\frac{\frac{1}{10}\cdot\frac{1}{\text{s}}\cdot\theta_\text{u}+\theta\left(0\right)}{\text{s}+\frac{1}{10}}\\ \\ \text{V}\left(\text{s}\right)=\frac{2\cdot\left(\Theta\left(\text{s}\right)-\frac{1}{\text{s}}\cdot\theta_\text{u}\right)+\text{v}\left(0\right)}{\text{s}+\frac{1}{2}} \end{cases}\tag3 $$
Now, distance is given by:
$$\text{v}\left(t\right)=\text{x}'\left(t\right)\space\to\space\text{V}\left(\text{s}\right)=\text{s}\cdot\text{X}\left(\text{s}\right)-\text{x}\left(0\right)\space\Longleftrightarrow\space\text{X}\left(\text{s}\right)=\frac{\text{V}\left(\text{s}\right)+\text{x}\left(0\right)}{\text{s}}\tag4$$
So, for the distance we get:
$$\begin{cases} \Theta\left(\text{s}\right)=\frac{\frac{1}{10}\cdot\frac{1}{\text{s}}\cdot\theta_\text{u}+\theta\left(0\right)}{\text{s}+\frac{1}{10}}\\ \\ \text{V}\left(\text{s}\right)=\frac{2\cdot\left(\Theta\left(\text{s}\right)-\frac{1}{\text{s}}\cdot\theta_\text{u}\right)+\text{v}\left(0\right)}{\text{s}+\frac{1}{2}}\\ \\ \text{X}\left(\text{s}\right)=\frac{\text{V}\left(\text{s}\right)+\text{x}\left(0\right)}{\text{s}} \end{cases}\tag5$$
Using Inverse Laplace transform:
$$\begin{cases} \theta\left(t\right)=\theta_\text{u}+e^{-\frac{t}{10}}\cdot\left(\theta\left(0\right)-\theta_\text{u}\right)\\ \\ \text{v}\left(t\right)=e^{-\frac{t}{2}}\cdot\left(5\theta_\text{u}-5e^\frac{2t}{5}\cdot\left(\theta_\text{u}-\theta\left(0\right)\right)-5\theta\left(0\right)+\text{v}\left(0\right)\right) \end{cases}\tag6$$
Finding the limit, when $t\to\infty$:
$$\lim_{t\to\infty}\text{x}\left(t\right)=\text{x}\left(0\right)+2\text{v}\left(0\right)+40\theta\left(0\right)-40\theta_\text{u}\tag7$$