Breaking up the force of the propeller into its large and small signal parts, we have $f_P = F_P + f_p$. The large signal parts give us $F_f = mg$. The small signal state equation is $\dot{x}$
$$\begin{equation*} \mathbf{\dot{x}} = \begin{bmatrix} \dot{v(t)} \\ \dot{h(t)} \end{bmatrix} = \begin{bmatrix} 0 && 0 \\ 1 && 0 \end{bmatrix} \begin{bmatrix} {v(t)} \\ {h(t)} \end{bmatrix} + \begin{bmatrix} 1/m \\ 0 \end{bmatrix} f_p(t) \end{equation*} $$First we find the step response for the open loop system from $f_P(t)$ to $h(t)$.
pkg load control
pkg load signal
clear all
m = 1 % Assume the mass
A = [[0, 0];[1, 0]]
B = [[1/m; 0]]
C = [0 1]
D = 0
sys = ss(A, B, C, D)
step(sys)
%g = place(sys, [-.1, -.2])
P = tf(sys)
%step(P)
Note that it blows up (but not exponentially). This is because with a constant force, we have a constant acceleration which means the height is going up as $t^2$. So it isn't good without feedback! This is what would happen if we had a unit step disturbance with no feedback.
Now, let's look at putting a controller in the system. We use the small signal model, and we want the small signal output for $h(t)$ to approach zero quickly. There can be a small signal disturbance and a small signal noise in our measurement of $h(t)$. These are incorporated into the model in blue below. (The black is the original system.) 
Let's see the disturbance transfer function step response for some possible controllers, $K(s)$. We want to make $K(s)$ large for low frequencies to make $G_d(s)$ small, but we want $K(s)$ small for large frequencies to make $G_n(s)$ small. So maybe a solution might be $K(s) = K_{big}/(s/\omega_0 + 1)$ or even $K_{big}/(s^2 +as + b)$. You need to be careful how many poles you put into $K(s)$ because each adds a phase delay, and too much delay is going to make the system unstable. A solution I like is the PD controller, where $K(s) = k_p + k_ds$.
K = tf([10000, 10000, 10000], [1, 0]) % For PD Controller (Looks like the D is really important for this system.)
%K = tf([10, 10], [1])
%K = tf(100, [1, 1])
%Gd = feedback(P,K)
%Gd = tf([1, 0],[1,0,10,0]) % Check the disturbance tronsfer function and make sure it isn't PK/(1+PK)
Gd = P/(1+P*K)
step(Gd)
You can play with $K(s)$ to see the effects. First we look at the Nyquist plot to see how far from -1, the loop gain is.
L = P*K
grid on
nyquist(L)
bode(L)
Gn = -P*K/(1+P*K)
bode(Gn)
Let's try to see what happens if we change the model a little by adding a viscous friction. So the system looks like this now: $$\begin{equation*} \mathbf{\dot{x}} = \begin{bmatrix} \dot{v(t)} \\ \dot{h(t)} \end{bmatrix} = \begin{bmatrix} -k/m && 0 \\ 1 && 0 \end{bmatrix} \begin{bmatrix} {v(t)} \\ {h(t)} \end{bmatrix} + \begin{bmatrix} 1/m \\ 0 \end{bmatrix} f_p(t) \end{equation*} $$
Let's try with $ 0.001< k < 0.1 $ N/(m/s), because it is supposed to be small.
k = 0.1
A = [[-k/m, 0];[1, 0]]
B = [[1/m; 0]]
C = [0 1]
D = 0
sys = ss(A, B, C, D)
%step(sys)
%g = place(sys, [-.1, -.2])
P = tf(sys)
step(P)
Note how the viscous friction makes a terminal velocity. Now the step response after the controller is installed shows a limited velocity.
Gd = P/(1+P*K)
step(Gd, 20)