Air resistance problem, expected time.

[emmaeng]

Homework Statement

As a bonus question on an assignment for my advanced physics class, we were asked to calculate (approx.) the expected time to impact of a drop of water falling 412m, from a tall building to the ground, considering linear drag. I've done most of the calculations but I expect that I'm missing something in the integrals. It's even more infuriating as they've given a "hint", and I still can't figure it out:

integral, from 0 to t of (1-e^-s/p)ds = t-p(e^-t/p)

which I worked out myself, so I can't quite figure out why they've given me this. Is it just that I can't rearrange this properly to give a value for t, or is there more to it?

where p is the time constant, p= V/g = 3.374 = m/c where c = 1.55x10^-7
V= terminal velocity = 33.07ms^-1
g= gravitational acceleration= 9.8ms^-2
m= 5.23x10^-7
s= 412m= distance traveled (vertical)

Homework Equations

with the above, I also have the equations of motion obviously as well as:
v = -V(1-e^-t/p), where v is the velocity at any given point (intially 0)

The Attempt at a Solution

So far all I've worked out is the terminal velocity and time constant (given above)
I think I need to use the height s (412m) and some relation between the time constant, terminal velocity and time but can't work through the maths.

Thanks heaps

 

[tiny-tim]

Welcome to PF!

Hi emmaeng! Welcome to PF!

I'm not sure why it's not working out for you, but it may be because this equation is wrong …

integral, from 0 to t of (1-e^-s/p)ds = t-p(e^-t/p)

if you put t = 0, you don't get 0, do you? ​