Falling Objects and Air Resistance

[galacticfan]

There are two objects, let's go with the classic cannon ball and football. These objects are both equal in size and shape but they obviously have different masses (the cannon ball being heavier). My question is, will these two objects fall and hit the ground at the same time if dropped from the same height on Earth and if their is drag (air resistance) acting upon them?

Some people that I have discussed this with say that momentum would make a difference and one would accelerate faster, whilst others say that the cannon ball would have more drag and so would travel at a slower speed but I have now become so lost in what to believe. I went back to basics and looked at this video: http://www.bbc.co.uk/learningzone/clips/gravity-momentum-and-air-resistance-on-falling-objects/10897.html but I thought that they should have hit the ground at the same time or was the result of that experiment just human error and the bottles needed to be falling at the same angle? Can someone please explain to me where I have gone wrong?

 

[Bandersnatch]

The force of gravity does impart all objects with the same acceleration equally, regardless of mass, so in the absence of any other force all objects will fall at the same rate.

The equation of motion in this case looks like this:
\vec{F}=\vec{F_g}
m\vec{a}=m\vec{g}

Where g is the gravitational acceleration vector, m is the object's mass.
Dividing by mass gives us the acceleration equation:

\vec{a}=\vec{g}

You can see that there's no mass anywhere in there. All objects accelerate with g.

However, with air present the drag forces appears, and the equation of motion looks like so:
\vec{F}=\vec{F_g}+\vec{F_d}
m\vec{a}=m\vec{g}-\frac{1}{2}ρ\vec{V^2}C_dA

Where in the drag force equation: ρ is the density of the fluid causing drag(air), V is the instantenous velocity, C_d is the drag coefficient depending on the shape of the body, and A is the cross-sectional area of the body.
Note that unlike the gravitational force, drag force is independent of mass.

Dividing by mass gives us the acceleration equation again:
\vec{a}=\vec{g}-\frac{1}{2m}ρ\vec{V^2}C_dA
In the case of two identical spheres(or whatever shape) of different masses, falling in the same conditions, ρ, A, and C_d are the same, but the drag acceleration(or deceleration - it's opposite to the velocity vector) increases with falling mass like 1/m, so the deceleration caused by drag increases for less massive bodies.

Note that bodies increase their velocity as they fall, and velocity increases drag as well, so given enough time they will attain the so called terminal velocity at which they don't accelerate any more. Terminal velocity is higher for more massive bodies(given all else is the same).

(edit: I had an artifact about integration there from my eariler draft. I hope it hadn't confused anyone.)

 

[galacticfan]

I'd like to thank you Bandersnatch for clearing that up and now it seems pretty obvious as to what the answer is.

 

[Bandersnatch]

Cheers.

To be more precise we should include buoyancy in the equation of motion for the case with air, but as long as the two bodies are relatively dense as compared to air, we shouldn't worry too much about that.

Anyway, I almost said welcome to the forum, but I see you've been around for a while, so I shan't.