# Homework Help: Terminal Speed and Height of Object

1. Sep 30, 2008

### Husker70

1. The problem statement, all variables and given/known data
(a) Calculate the terminal speed of a wooden sphere(density .830 g/cm^3)
falling through the air if its radius is 8.00cm and its drag coefficient is .500.
(b) From what height would a freely falling object reach this speed in the
absence of air resistance?

2. Relevant equations
p =m/v
m=pv
Vt = sqrrt 2mg/DpA

3. The attempt at a solution
p = .830 g/cm3
r = 8.00 cm = .080m
A = pie(r2) = .020m2
D = .500

Volume of Sphere = 4/3 pie(r3) = .335m3

I have the Velocity equation ready to go but I don't think that I'm converting
p correctly. I get .830 g/cm3 = 8.3x10 -7 g/m3

I'm not sure how to go to kg/m3, which is what I believe that I need to find
speed since g = m/s2
Once I get p to be correct I can get m and p and put them into the Vt equation.
I think I'm confused on the conversion.
Thanks,
Kevin
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 30, 2008

### PhanthomJay

The conversion factor is 1g/cm^3 = 1000kg/m^3. The density of the wooden sphere may be used to calculate the mass of the sphere, but that is not the correct density to use in your equation for V_term.

3. Sep 30, 2008

### Husker70

ok so .830 g/cm3^3 = 830kg/m^3
so
m=pv = (830kg/m^3)(.335m^3) = 278kg
Is that right?
Thanks,
Kevin

4. Sep 30, 2008

### Husker70

Ooops I get Volume of sphere to be .0021m^3
So m=pv = 830kg/m^3(.0021m^3) = 1.74kg

So I get Vt to be 4.11m/s
Can anyone help confirm this.
Thanks,
Kevin

5. Sep 30, 2008

### PhanthomJay

The density looks good, but the volume is (4/3)(pi)(r^3) =4/3(pi)(.08m)^3= .002m^3, no? Now after you get the density and volume of the wood sphere correct (it's not that easy with all those metric decimals), calculate the mass of the sphere. But i want to be sure you understand that when you calculate V-term, the density to use in the denominator of that equation is NOT the density of the wood, it is the density of the ____?

6. Sep 30, 2008

### Husker70

I still get .0021m^3 for volume. Maybe I don't understand that.
In the Vt don't I have to find the mass of the sphere and use that?
Kevin

7. Sep 30, 2008

### PhanthomJay

I said .002 which is just a round off of .0021; your number is OK. But you're missing the point. When you calculate the terminal velocity using your relevant equation for V_t = sqrt2mg/DpA, the number to use for p is the density of the air, not the wood.

8. Sep 30, 2008

### Husker70

That does make sense. Sorry I misunderstood.
So would I use 1.2kgm^3 or do I have to calculate it from this
problem?
Kevin

9. Sep 30, 2008

### Husker70

As the density of air

10. Oct 1, 2008

### PhanthomJay

You use the 1.2kg/m^3 which you have looked up and correctly identified as the approximate density of air near earth's surface. Now plug in all the numbers to solve for the terminal velocity.