Upthrust vs Drag 
24: A ball is falling at terminal speed in still air. The forces acting on it are upthrust, viscous drag and weight. What is the order of increasing magnitude of these 3 forces?

Answer :upthrust  < viscous drag  < weight

Could some one please explain why in still air upthrust of a ball is smaller than the viscous drag ?
 A: Although it isn't absolutely clear from the question I would guess that the up thrust is the force due to Archimedes' principle, so the up thrust is equal to the weight of air displaced by the object. Since the object is not accelerating, the net force on it must be zero so:
$$ mg = \text{drag} + \text{up thrust} $$
Obvious both the drag and up thrust must be less than or equal to $mg$, but their relative size will depend on the density of the ball. You would have to do the calculation to determine which was bigger. Unless your ball is something exotic, like a hot air balloon, a back of the envelope calculation should make it pretty obvious that the up thrust is a tiny force.
A: Up thrust is approximately constant:
$$U=\rho\cdot g\cdot H$$


*

*At $H$ height, air density low = up thrust

*At $H-x$ (when object travel certain height), air density increase=up thrust


Comparing statements 1 and 2, up thrust is constant at $H$ height and $H-x$
BUT
drag or viscous force increases as velocity of an object increases,
so viscous drag is greater than up thrust.
