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In the case of a free fall object, if $h$ denotes distance from floor and $v$ the speed, then, given an input $U = g$, and $X=[h \quad v]^T$ and $\dot{X}=[\dot{h} \quad \dot{v}]^T$, I found that the state space equation of $B$ contains $t$, that is $B=[t \quad 1]^T$. I used $\frac{dh}{dt}=v + gt$ to get that result.
Does it mean that this system is not time-invariant but in fact, time variant? Am I correct, or did I do something wrong?
This is the model that I follow.
In your notation you are mixing up the integration of v. For the state space representation one would write: $$\begin{pmatrix} \dot h \\ \dot v \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} h \\ v \end{pmatrix} + \begin{pmatrix} 0 \\ \frac{1}{m} \end{pmatrix} F_\text{ext}.$$ This expresses the equations of motion of a free point mass $m$ in one dimension $h$ with external force $F_\text{ext}=mg$. The change of position / height is $\dot h(t) = v(t)$ and the change of velocity depends on the force: $\dot v = \frac{1}{m} F_\text{ext} = g.$ In the state-space terminology we have the state vector $x=(h,v)$ and the force is the input of the system. (Even for time-dependent $F_\text{ext}(t)$ the system is time-invariant.)
Writing $\dot h = v + g t$ is not correct. We have $\dot v(t) = g$ which integrates to $v(t) = v_0 + g t$ with integration constant $v_0$. Integrating again gives $\dot h = h_0 + v_0 t + \tfrac 1 2 g t^2$ with integration constant $h(t=0)=h_0$.
Supplement to answer comment: For a drag force that is proportional to velocity $F_\text{drag}=-b\, v(t)$, the velocity's equation of motion changes to $\dot v=g - \frac{b}{m} v$. To express this, one would add a constant matrix element to the system function, so the system is still time-invariant: $$\begin{pmatrix} \dot h \\ \dot v \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & \frac{-b}{m} \end{pmatrix} \begin{pmatrix} h \\ v \end{pmatrix} + \begin{pmatrix} 0 \\ \frac{1}{m} \end{pmatrix} F_\text{ext}.$$
