Fall with drag - ignoring the gravitational force? Say we have a fall while considering air resistance
$$ma=mg-Bv^2.$$
It is a standard differential equation.
But then I was asked what would happen, if we ignore the gravitational force $mg$. Now I am confused, why or when can we ignore the gravitational force? I assume when the object is very very small so that it doesn't matter anymore? But then
$$m\frac{d^2 s}{dt^2}=-B\left(\frac{ds}{dt}\right)^2.$$
What exactly does this say about the situation?
 A: @MarkoGulin has fully addressed the mathematical aspect of the problem. I thus add a remark regarding the physical situations where the equation in question can be relevant:

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*Drag force is ubiquitous when we talk about movement of macroscopic objects in a gas or liquid. It is not hard to think of examples that do not involve free fall: e.g.,

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*a boat or a submarine that has turned off its engines,



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*a glider aircraft,



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*an immobile helicopter or a baloon carried by a gust of wind,



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*a meteorite or a satellite entering the Earth's atmosphere, etc.



*Projectile motion: here we are dealing with a bit more complicated situation, since the drag force is proportional to the square of the full velocity:
$$m\dot{\mathbf{v}}=m\mathbf{g}-B\frac{\mathbf{v}}{|\mathbf{v}|}\mathbf{v}^2.$$
However, when the horizontal velocity is very big, we can approximate it by
$$m\dot{v}_x=-Bv_x^2.
$$
A: This is a simplification when drag is much larger than the gravitational force. In order for this to work the drag must be at least order of magnitude larger. When initial velocity is (close to) zero, the simplification obviously does not work as it predicts the object to remain at rest.
The free-fall example is little bit difficult to imagine. Much better example would be a moving car with no force from the motor, assuming the car moves on a horizontal surface (road).

If the equation of motion is given as:
$$m \frac{d^2}{dt^2} x(t) = -k \cdot v(t)^2$$
it can be written as
$$m \frac{d}{dt} v(t) = -k \cdot v(t)^2$$
which equals to
$$\int \frac{dv(t)}{v(t)^2} = -\int \frac{k}{m} dt$$
This finally leads to
$$\frac{1}{v(t)} = \kappa t + C_1$$
where $\kappa = \frac{k}{m}$, $C_1 = \frac{1}{v_0}$, and $v_0$ is the initial velocity.
The equation for velocity is
$$\boxed{v(t) = \frac{v_0}{1 + \kappa v_0 t}}$$
From this it is trivial to find the equation for position:
$$\boxed{x(t) = \frac{1}{\kappa} \ln(1 + \kappa v_0 t)}$$
Note that the velocity approaches zero with time, i.e. the drag is slowing the object.
