A question of terminal velocity!

We know a little about gravity. And the increase in velocity of an object when dropped from a height, say from an airplane.

Since g = 32 ft/s2, the velocity will be 32 ft/s (9.8 m/s) after one second.

Fine. Now we have something called 'terminal velocity'.

so, will a penny dropped from an aircraft fall faster than a baseball?

Originally posted by mikesingh
so, will a penny dropped from an aircraft fall faster than a baseball?

No, I'd imagine the baseball would fall faster due to its more aerodynamic shape. The penny would most likely be pushed and thrown around by the wind, slowing its fall. Though it's more dense than the baseball, it's less massive, meaning the baseball will accelerate more as it falls.

So to answer your question, the baseball would have higher terminal velocity than the penny.

Terminal velocity is dependent on the cross section of the object and its mass.

Assuming a coin has a crossection of 2cms, thats 0.0004m^2 and a weight of 0.009kgs, the baseball 0.14Kgs and 0.05m^2.

Assuming a drag of 1 and an atmospheric density of 1

v of coin = 21
v of baseball = 7.5

so the coin is moving at around 3 times that of the baseball at terminal velocity.

this page has more info
en.wikipedia.org...

[edit on 3-8-2006 by CIS001]

[edit on 3-8-2006 by CIS001]

I think thousand's got it closer - a penny would probably approach terminal velocity faster at first due to it's lower drag characteristic, but turbulence would end end up kicking it about all over the place, and the baseball would over take due to it's more aerodynamic shape...

Angel42

Originally posted by CIS001
Terminal velocity is dependent on the cross section of the object and its mass.

Assuming a coin has a crossection of 2cms, thats 0.0004m^2 and a weight of 0.009kgs, the baseball 0.14Kgs and 0.05m^2.

Assuming a drag of 1 and an atmospheric density of 1

v of coin = 21
v of baseball = 7.5

so the coin is moving at around 3 times that of the baseball at terminal velocity.

It's a bit closer together than that. Ignoring skin drag on the ball (which is a safe assumption for now) you get (using your masses and cross-sections)

v of coin = 17 m/s
v of baseball = 10 m/s

because you can't assume "Cd = 1.0" for both. The Cd for a circular flat plate (the penny) is 1.12 and the Cd for a sphere (baseball) is .47.

Originally posted by angel42
I think thousand's got it closer - a penny would probably approach terminal velocity faster at first due to it's lower drag characteristic, but turbulence would end end up kicking it about all over the place, and the baseball would over take due to it's more aerodynamic shape...

Angel42

The penny is actually more draggy, and also doesn't have the force acting downward to accelerate as fast as the baseball, so it actually takes the penny longer to reach terminal velocity than it does the baseball.

baseball reaches terminal velocity in about 2 to 3 seconds of being dropped
penny reaches terminal velocity in about 5 to 6 seconds.

EDIT: Re-worded something that didn't seem to read right to me.

[edit on 8-3-2006 by Valhall]

Thankyou, I had a feeling the assumptions I made wouldn't quite work

Of course, both of us figured the coin falling face first...which probably isn't correct, but for the sake of comparing the two shapes, we'll pretend it would.

If it went on edge (which I'm assuming it most likely would since that's the profile of least resistance) - all bets are off...lol. (What's the thickness of a penny?)

We're going to get this to a gnats-butt. Hyperphysics has baseball radius at 3.66 cm, Area at .0042 m^2 and a mass of .145 kg. They list the terminal velocity at 32 m/s and when I put their mass and cross-section into my spreadsheet, I get 34 m/s.

Now, we need to do the penny on it's edge.

[edit on 8-3-2006 by Valhall]

OK guys,
Here's the answer... (under the heading 'Terminal Velocity')

www.school-for-champions.com...

Have a nice day!