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I was conversing with my professor about the forces acting on a ball that is thrown onto the air. He said that you would have the weight of the ball, air resistance and buoyant force. Although, I wonder if the buoyant force has a significant impact on the result, assuming that the ball's density is considerably greater than that of the air.

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  • $\begingroup$ It of course depends on what a ball is. The difference between a balloon, a volleyball and a ping-pong ball is just the thickness of the plastic shell. $\endgroup$
    – MaxW
    Oct 5, 2020 at 16:47

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Yes, the ball does experience an upward buoyancy force equal to the weight of the displaced air - not just when it is thrown, but when it is on the ground too. However, since the density of air is about $\frac 1 {1000}$ that of water, the effect of atmospheric buoyancy on the apparent weight of a solid ball is negligible. It is about an order of magnitude less than the effect of the earth's rotation. In other words, if you are not adjusting $g$ to account for your latitude then you don't need to worry about atmospheric buoyancy either.

All of the above assumes that the ball has a density much greater than that of air. Obviously there are some objects with low average density, such as balloons, for which atmospheric buoyancy is significant.

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Although, I wonder if the buoyant force has a significant impact on the result, assuming that the ball's density is considerably greater than that of the air.

Not sure what "result" you are referring to. But if the question is will the ball continue to rise after release, then the answer is only if the upward buoyant force is greater than or equal to the sum of the weight of the ball and air resistance or drag force, $F_{D}$, opposing its motion.

$$ρVg≥mg+F_{D}$$

The left side is the upward buoyant force where

$ρ$= the density of the air at the location of the ball

$V$= the volume of air displaced by the ball = volume of the ball

$g$= acceleration due to gravity.

$mg$ is the weight of the ball

$F_{D}$ is the drag force, which is proportional to the square of the relative velocity between the ball and surrounding air.

In order for the buoyant force to be greater than weight of the ball, the density of the ball (mass/volume) has to be less than the density of surrounding air. That, in turn, means most of the volume of the ball would need to consist of a gas lighter than air (e.g. helium).

If the ball density is considerably greater than the air, the impact of the buoyant force would be insignificant.

Hope this helps

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  • $\begingroup$ This does not address the question. The ball is not rising due to buoyant force. $\endgroup$
    – nasu
    Oct 5, 2020 at 18:10
  • $\begingroup$ Your question is vague. You asked "Although, I wonder if the buoyant force has a significant impact on the result" but you didn't say what result you were wondering about. I assumed you wanted to know whether or not the ball will continue to rise after being released. I'm trying to say the ball will only continue to rise if the upward buoyant force is greater than the weight of the ball. I will update, though I suppose it is moot since you already accepted an answer. $\endgroup$
    – Bob D
    Oct 5, 2020 at 20:11
  • $\begingroup$ I am not the OP. $\endgroup$
    – nasu
    Oct 6, 2020 at 11:22
  • $\begingroup$ Sorry, thought you were $\endgroup$
    – Bob D
    Oct 6, 2020 at 11:33

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