Ball and air resistance (curiously difficult)

[nicholaslyz]

Hi,

I was wondering in my physics class the other day, whether if you throw up a ball from a height say h with velocity v₀ , facing air resistance F_air=-bv on its way up and down, would it reach its initial height h again faster or slower than if you were to throw it up in a vacuum (i.e. F_air=0)?

It's hideously complicated...for starters the time taken for the ball to reach its maximum height in the presence of air resistance is [tex]t =- \frac{m}{b}\ln \frac{1}{1+\frac{bv_0}{gm}}[/tex]

You can check this by [tex]\lim_{b \to 0}\Big(\textstyle \frac{-m}{b} \ln \frac{1}{1+\frac{bv_0}{gm}}\Big)=\frac{v_0}{g}[/tex] , so if b = 0 then the equation is the one for time taken to reach the maximum height in vacuum.

So does the ball rise and fall faster in air resistance?

 

[Naty1]

How would say a steel ball behave different in molasses??

It's a frictional force...analogous to frictional drag on an inclided plane...

Am unsure what you imply by "faster"...
the object will slow down faster as it initially rises because it is doing work (losing kinetic energy) against gravity AND air resistance. so it will not rise as far...then it falls more slowly due to air drag...and might reach some terminal velocity...

I don't intuitively see whther it will remain airbore for a different period of time or not.

I check projectile trajectory formulas...for example, a horizontally fired projectile falls according to vertical d = 1/2gt² when ignoring air resistance; there should be some standard formula around for the effects of air drag...maybe that will help.