dropping balls
 

If you were to drop two racquetballs, one filled with lead and the
other normal, off of the Sears Tower, which one would hit the ground
first?

A) neglecting air resistance
B) including air resistance
C) Does the answer change for either A or B if one ball is many orders of magnitude heavier then the other.

I was wrong the first time I tried to answer this question, and it spawned a 100+ post thread at another forum.

Re: dropping balls
 

i say they drop the same time...

Re: dropping balls
 

Didn't Galileo already do this?







I thought I saw him at the Sears Tower..... :p

Re: dropping balls
 

A) both hit at same time
B) the one filled with lead
C) no

Re: dropping balls
 

Hmm... I'd say they never hit the floor. They would get stuck midway. Of course neglecting air resistance I would say they would hit at the same time.

Re: dropping balls
 

a. the same
b. the same
c. no


They are the same shape and therefore have the same terminal velocity.

Andy B.

Re: dropping balls
 

Sorry, Andy, but terminal velocity also depends on density. A lead-filled sphere will have a higher terminal velocity than an air-filled one. Try dropping a ping-pong ball and a rubber ball of the same size -- the ping-pong ball reaches terminal velocity very quickly, but the rubber ball keeps accelerating and outpaces it.

The answer to the question is that air resistance matters when the two objects' densities are different. A heavier ball having the same size as a lighter one will hit the ground first, if you drop them from high enough for the lighter one to reach terminal velocity.

Re: dropping balls
 

1. Neglecting air resistance, they will reach the ground at the same time.

2. The force of drag on an object is described by the following equation:

F = 0.5*Cd*rho*A*V^2 (eq 1)

where:

Cd = coefficient of drag
rho = density of air
A = frontal area of the object
V = airspeed of the object

The terminal velocity of the object is when its weight (mass * accel due to gravity) is equal to the force of drag. Thus, the equation becomes:

m*g = 0.5*Cd*rho*A*V^2 (eq 2)

We can then solve eq 2 for V to get the terminal velocity:

V = sqrt[(2*m*g)/(Cd*rho*A)]

Since the two raquetballs are identical except for the mass, the Cd and A are the same for the two balls. Therefore, as the mass increases, the terminal velocity of the ball increases. Hence, the lead filled ball hits the ground first. However, the terminal velocity increases proportionally to the SQUARE ROOT of the mass, (i.e. not directly proportional to the mass).

3. The answers are the same. It doesn't matter for case 1. For case 2, the terminal velocity still increases as the square root of the mass increases.

Re: dropping balls
 

Thanks, Alan... it is a good thing that I'm not designing airplanes!

I'm glad to be on the same team as smart people.


Andy B.

Re: dropping balls
 

Can you link us to this 100 post thread on the other forum? ;)

Re: dropping balls
 

Hmm...I'm designing airplanes (careful if any of you have the opportunity to fly in a plane that I have worked on...though you'd have to be in the Navy)...does that mean I should be able to answer this question? :D


Well, I don't have my aero books on me (I left them at school) but all those equations Chris was spouting out sure does look familiar. So, lets see...if I had to answer them, my answers would be:

A) the same
B) the lead filled ball
C) no, they don't change

Re: dropping balls
 

well, I would say A) the same time B) the lead one first and C) no

BUT, it doesn't say that they are dropped at the same time, so it could be A) which ever you dropped first B) which ever you dropped first unless you dropped the lead filled almost instantly after the normal and C) would remain the same.

but then I was thinking... is there a point on the tower which you can drop something so that it will go straight down to the ground?

Re: dropping balls
 

All of you are neglecting one thing and that's the fact that there should be pretty good cross winds when you drop the balls due to the effects of being ina city. This is the same reason why pennys never reach the bottom of the Empire State building. Im not sure how powerful these winds are but they might affect the outcome.

Re: dropping balls
 

cross winds would have a more dramatic effect on the lighter of the two balls, and therefore wouldn't affect which one would reach the ground first. now if there was a powerful enough downdraft, and it was a relativly short drop, maybe the lighter of the two would hit the ground first.

Re: dropping balls
 

Who wants to go to Chicago and test this? ;)

I'm pretty sure that it is
a) Same
b) Lead
c) No

Although B may be wrong, I don't know if the air resistance varies due to density and weight. If they don't factor, then it's the same because both balls have the same size, shape, and surface.