Well. A Squared, you'd be wrong.
After your first response I went and paid a visit to a young lady (physics grad) who tutored my daughter in Physics a couple of years back.
S = ut + 1/2 at^
^ = squared
She said this is the valid equation up until the point the object reaches Vt, as the object is still accelerating and mass does not come into it, so in a fall where the objects fail to reach their Vt the above equation is it and they will hit the ground at the same time. Now, if the distance the objects fall is such that Vt will be reach prior to their hitting the ground the Vt equation of
will come into it. At this point mass impacts Vt and therefor the higher mass object will have a higher Vt and will hit the ground earlier. So the as the lighter object reaches Vt (now falls at its Vt as determined by the Vt equation) it will now stop accelerating while the higher mass object continues to accelerate (using s =ut +1/2 at^), until it reaches Vt.
So as to the effect of mass depends on whether the object hits the ground prior to reaching Vt.
She laughed at your helium balloon comparison, if you check a reasonable link the Vt equation like at this link
Terminal velocity - Wikipedia, the free encyclopedia
you'll see it mentions where buoyancy isn't a consideration.
Mass comes into the Vt equation as it's "pushing" against the air resistance, thus a higher mass object will have a higher Vt.