The terminal velocity of a falling body is attained when its weight is balanced by the drag force.

Let’s assume that the drag force is proportional to the velocity of the falling body. So we write the drag force as kv, where k is a constant that is dependent on how streamline the falling body is.

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We can then solve for the terminal velocity v_{t} for the single sheet of paper dropped horizontally:

kv_{t} = mg

v_{t} = mg/k

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When a stack of papers are dropped (instead of a single sheet), k is unchanged (since it is the same area pushing into air), but m is of course multiplied by the number of sheets of paper in the sheet. For a stack of 20 sheets, the terminal velocity would be increased by 20 times.

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For the single sheet dropped vertically, m is unchanged, but k is reduced dramatically (since now only a tiny area is pushing against air). Assuming the k is directly proportional to the cross-sectional paper, k would be reduced by more than 1000 times (because the cross-sectional area is reduced from 216 mm x 297 mm to about 0.05 mm x 297 mm). This implies that the terminal velocity would be increased by more than 1000 times.