# Thread: falling object with air resistance

1. ## falling object with air resistance

(eight-pound object dropped from a height of 5000 feet, where air resistance is proportional to the velocity.)

write the velocity as a function of time if its velocity after 5 secs is approximately -101 feet per second.

2. Originally Posted by AAA
(eight-pound object dropped from a height of 5000 feet, where air resistance is proportional to the velocity.)

write the velocity as a function of time if its velocity after 5 secs is approximately -101 feet per second.
F= ma so (8/32)a= (1/4)dv/dt= -32+ kv. v(0)= 0. Can you solve that equation for v? It will contain a k of course so you need to use v(5)= -101 to find k.

3. Originally Posted by HallsofIvy
F= ma so (8/32)a= (1/4)dv/dt= -32+ kv. v(0)= 0. Can you solve that equation for v? It will contain a k of course so you need to use v(5)= -101 to find k.

Thanks, but why 8/32?

4. F = mg

8 lbs is the Force due to gravity.

8 = m(32) ... acceleration due to gravity is 32 ft/s^2

m = 8/32 = 1/4 slug

yes, slugs ... one of the joys of the English system of measurement

5. ## Object falling with air resistance

Originally Posted by skeeter
... yes, slugs ... one of the joys of the English system of measurement
True, but don't blame us - we use SI units, and have done for 40 years or so! (All right, they are a bit French, but we all have to swallow our pride at some time or other!)

In fact, slugs are horrible creatures, best avoided. Just call the mass $\displaystyle m$ - it cancels anyway.

The equation of motion is then:

$\displaystyle -mg + k'v = m\frac{dv}{dt}$, where $\displaystyle v$ is measured vertically upwards

And you can then divide through by $\displaystyle m$, and replace $\displaystyle \frac{k'}{m}$ by $\displaystyle k$ to obtain:

$\displaystyle -g + kv = \frac{dv}{dt}$

After separating the variables, integrating, and using $\displaystyle v(0) = 0$, this gives:

$\displaystyle v = \frac{g}{k}(1 - e^{kt})$

So when $\displaystyle t = 5, v=-101$ (and $\displaystyle g = 32$):

$\displaystyle -101 = \frac{32}{k}(1 - e^{5k})$

How you are expected to solve this equation for $\displaystyle k$, I don't know (Newton's method?). In fact, I found $\displaystyle k \approx -0.2$ (but I cheated and used a spreadsheet!)

So the solution is

$\displaystyle v\approx 160(e^{-0.2t}-1)$