1) ![161 \frac{lbf \cdot s}{ft}](https://tex.z-dn.net/?f=161%20%5Cfrac%7Blbf%20%5Ccdot%20s%7D%7Bft%7D)
The particle experiences two forces during its fall:
- The force of gravity,
, pointing downward, where m is the mass of the particle and g is the acceleration of gravity
- The drag force,
, pointing upward, where k is a constant and v is the instantaneous speed of the particle
When the particle reaches its terminal speed, the acceleration becomes zero: this means that the net force is also zero, so the two forces are balanced. Therefore we can write
![mg = kv_t](https://tex.z-dn.net/?f=mg%20%3D%20kv_t)
where
is the terminal velocity. We have the following data:
is the mass
is the acceleration
is the terminal speed
Solving for k, we find
![k=\frac{mg}{v_t}=\frac{(1 \frac{lbf \cdot s^2}{ft})(32.2 \frac{ft}{s^2})}{0.2 \frac{ft}{s}}=161 \frac{lbf \cdot s}{ft}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bmg%7D%7Bv_t%7D%3D%5Cfrac%7B%281%20%5Cfrac%7Blbf%20%5Ccdot%20s%5E2%7D%7Bft%7D%29%2832.2%20%5Cfrac%7Bft%7D%7Bs%5E2%7D%29%7D%7B0.2%20%5Cfrac%7Bft%7D%7Bs%7D%7D%3D161%20%5Cfrac%7Blbf%20%5Ccdot%20s%7D%7Bft%7D)
2) 28.6 ms
We now want to find the time at which the particle reaches 99 % of the terminal speed, so a speed of
![v' = 0.99 v_t = (0.99)(0.2)=0.198 ft/s](https://tex.z-dn.net/?f=v%27%20%3D%200.99%20v_t%20%3D%20%280.99%29%280.2%29%3D0.198%20ft%2Fs)
The net force acting on the particle is
![F=mg-kv](https://tex.z-dn.net/?f=F%3Dmg-kv)
So the acceleration is
![a=\frac{dv}{dt}=g-\frac{k}{m}v](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bdv%7D%7Bdt%7D%3Dg-%5Cfrac%7Bk%7D%7Bm%7Dv)
This differential equation can be rewritten as
![\frac{dv}{g-\frac{k}{m}v}=dt](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bg-%5Cfrac%7Bk%7D%7Bm%7Dv%7D%3Ddt)
And by integrating on both sides and performing the calculation, we can get the final expression for the velocity as
![v=\frac{mg}{k}(1-e^{-\frac{k}{m}t})](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bmg%7D%7Bk%7D%281-e%5E%7B-%5Cfrac%7Bk%7D%7Bm%7Dt%7D%29)
We want to find the time t at which v = 0.198 ft/s, so we now need to re-arrange the formula for t. This gives:
![t = -\frac{m}{k} ln (\frac{g-\frac{k}{m}v}{g})](https://tex.z-dn.net/?f=t%20%3D%20-%5Cfrac%7Bm%7D%7Bk%7D%20ln%20%28%5Cfrac%7Bg-%5Cfrac%7Bk%7D%7Bm%7Dv%7D%7Bg%7D%29)
And substituting all the values, we find:
![t = -\frac{1}{161} ln (\frac{32.2-\frac{161}{1}(0.198)}{32.2})=0.0286 s = 28.6 ms](https://tex.z-dn.net/?f=t%20%3D%20-%5Cfrac%7B1%7D%7B161%7D%20ln%20%28%5Cfrac%7B32.2-%5Cfrac%7B161%7D%7B1%7D%280.198%29%7D%7B32.2%7D%29%3D0.0286%20s%20%3D%2028.6%20ms)