Question

2 hailstones of radii are in ratio of 1:2 fall from a height of 50km. their terminal velocities are in what ratio

Answer 1

Equation of motion for a body falling from a height in presence of air (fluid with buoyancy and drag force).. 
    y = displacement,   v = velocity, b = drag factor/coefficient
    d_ice = density of ice,  d_air = density of air
    m = mass of hailstone = 4π/3 * R³ * d_ice
    Buoyancy force =  mg * d_air/d_ice = 4π/3 R³ * d_air * g
    drag force = - b v

  m dv/dt = mg [1 - d_air/d_ice] - b v
  dv/dt = g [1 - d_air / d_ice]  -  b/m * v
        =  0 for terminal velocity

So  v_t = g [1 - d_air /d_ice ] * m/b
              = 4π/3 r³ [d_ice - d_air] / b

Terminal velocity is proportional to mass , volume or cube of radius of the hailstone.

Ratio of radii = 1 : 2.
Ratio of terminal velocities = 1 : 8