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Dropping two balls of identical circumference from the same height always has resulted...
Dropping two balls of identical circumference from the same height always has resulted in the heavier one hitting the ground first. This is not what was expected as they should, if the only two forces are gravity and air resistance, hit the ground at the same time. Any ideas what is going on here. Is there another force in play (bouyancy?).
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(Level 1) Associate Educator, Expert
First of all, let me clarify that we expect that the two objects will hit the ground at the same time if we neglect air resistance. This is so because, acceleration, without air resistance, is independent of the mass of the falling object.
An object falls because of gravity, and the acceleration due to gravity is usually assumed to be constant (depending on which planet you are considering). This constant is 9.8 m/s^2. Now, the expression for velocity is:
`v_f = v_i + at` and it is not dependent on the mass of the object.
In reality, however, this is not what we observe. A heavier ball will hit the ground before a lighter one, just as a billiard ball or a hammer, will hit the ground before a feather.
This is so because in reality, there is air resistance, and drag. The assumption that only gravity is acting on the object is not accurate in the real world, as there are a lot of other factors -- again, air resistance, drag, wind, etc.
Buoyancy of air, is the same principle as that of buoyancy in water. Buoyancy is an upward force (push) of air or water on a submerged object. The volume of air/water displaced is related to the buoyant force. The balls in question here have the same volume (same circumference) but one is heavier (and is hence denser). Taking into consideration buoyancy, the heavier ball will fall first (derivation of a formula [see video in reference] shows that `a = g(1- (rho_(air))/(rho))` where a is the acceleration of the object, and g the constant, while rho is the density of the object.) Hence, buoyance is also a factor including drag/air resistance. ``
Posted by mvcdc on October 2, 2013 at 10:45 AM (Answer #1)
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