Change in appearance of liquid drop due to gravity A liquid drop is spherical in shape due to surface tension. But why does it appear as a vertical line under the free-fall due to gravity? (E.g. During a rain - falling raindrop) Is there a specified length for the line or does it vary with the size of drops?
 A: A drop that is free falling in vacuum is spherical. This is because free falling in a gravitational field is the same thing as being at rest with no gravitational field present: the gravitational field and the acceleration cancel each other out.
Rain drops falling to the earth can have various shapes depending on their size, although I am not aware that they can become elongated (can you provide a source?). These shapes are due to the air flowing past them, in a rather intuitive way: roughly, air flow causes the bottom to become flatter.
Edit: Regarding the appearance of raindrops (as opposed to their physical shape), consider taking a photograph of falling rain. The camera integrates the incoming light over the exposure time $t$, during which the drop travels a distance of $v t$, where $v$ is the velocity. If we are close to the ground this will be the terminal velocity, which is about $2_{m/s}$. If we use an exposure time of $t=1/60_s$ (say we are using a flash), the drop will trace a line of length $\sim 3_{cm}$. The apparent line on the photograph then has to account for distance from the drop, etc.
A: It is not an answer but an unfinished draft of how to get the shape :
Let $y(x)$ be the profile function of this axisymmetric shape. Supposing a laminar regime friction implies the total counterforce to weight is : $$F=2\pi av\int_0^L\frac{y'y}{\sqrt{1+y'^2}}dx$$ because the friction force is along the normal, while the projection along the axis is needed.
This is a variational problem, the minimum of F is seeked under the constraint of constant volume, given by : $$V=\pi\int_0^L y^2dx$$ and $y(0)=y(L)=0$.
Using Lagrange multipliers we get the functional : $$I[y]=\int_0^L \frac{\pi a v y y'}{\sqrt{1+y'^2}}-\lambda( \frac{V}{L}-\pi y^2)dx$$. The Euler-Lagrange equation permits to express y as a function of the multiplier and the derivatives, but I stopped there because it becomes a bit tedious and it is late here.
Then should follow to derivate towards $\lambda$ getting a final ODE for y, nevertheless it is nonlinear and of high order.
The latter should imply in particular that there are families of shapes, thus maybe there is no agreement on it.
