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Answer: depending on viscosity, mass and velocity of impact, if droplet integrity is maintained, as velocity increases, diameter will increase from
Approximately d*sqrt(2) to 2*sqrt(d^3/6a) where d is original diameter and a is thickness when the droplet flattens into a disc.
Explanation:
This applies generally to any liquid droplet, which by inference falls and impacts a solid surface. The impact force is mgh where m= mass, g= acceleration due to gravity, h= initial height.
A liquid droplet deforms on impact. Assume the drop is sperical, then the deformation distance, d= diameter of the droplet then the average impact force = mgh/d.
The droplet may spread, splash or bounce, depending on viscosity and force, which depends on mass and velocity immediately before impact.
All than can be said is that if the droplet maintains integrity it could achieve the shape of half a highly flattened oblate spheroid. Approximating this with a flattened disc of thickness a, and an original volume of 4/3pi(d/2)^3, the volume as a disc =a*pi*r^2 so the horizontal diameter = 2*sqrt(d^3/6a)
It is not really possible from the available data to determine whether the droplet would remain its integrity, but at sufficiently low force/velocity, the droplet could retain a near-hemispherical shape, giving a horizontal diameter of the hemisphere = d*sqrt(2)
As velocity increases, if integrity is maintained, the diameter will increase from the second approximation to the first