Solution

Let's consider a small part of the cobra with length $\Delta x$ and mass $\Delta m$. When this part is lifted off the ground, it acquires a momentum:
$$\Delta P = \Delta m \cdot v$$

Since the snake is homogeneous, the mass of this element is:
$$\Delta m = \frac{m}{l} \Delta x$$

By Newton's second law, the additional force associated with accelerating these mass elements from rest to speed $v$ is equal to the rate of change of momentum:
$$F = \frac{\Delta P}{\Delta t} = \frac{\Delta m}{\Delta t} v = \frac{m}{l} \frac{\Delta x}{\Delta t} v$$

By the definition of speed, $\frac{\Delta x}{\Delta t} = v$. Whence the force associated with the acceleration of the mass is:
$$F = \frac{mv^2}{l}$$

Let's take everything together and find the total support reaction force $N$. The ground must support the entire gravitational weight of the snake ($mg$) plus provide the extra force $F$ to accelerate the lifting parts:
$$N = mg + \frac{mv^2}{l} = m\left(g + \frac{v^2}{l}\right)$$

By Newton's third law, the force with which the cobra presses on the ground is equal in magnitude to the support reaction force $N$.