Question

You dive straight down into a pool of water. you hit the water with a speed of 6.5 m/s , and your mass is 70 kg . part a assuming a drag force of the form fd = (−1.10×104 kg/s ) v, how long does it take you to reach 2% of your original speed? (ignore any effects of buoyancy.)

Answer 1

The differential equation would be:

$m \frac{dv}{dt} = -bv-mg$

To solve I'm going to use the integration factor method with:
m mass
b = 1.1 * 10⁴ kg/s
g = 9.81 m/s²
t time
$integration factor = e^{\int\limits \frac{b}{m} \, dt} = e^{ \frac{b}{m} t}$

$\frac{dv}{dt} + \frac{b}{m}v = -g \\ \frac{dv}{dt} e^{ \frac{b}{m} t} + \frac{b}{m} e^{ \frac{b}{m} t}v = -g e^{ \frac{b}{m} t} \\ ve^{ \frac{b}{m} t} = \int\limits {-ge^{ \frac{b}{m} t}} \, dt = -g \frac{m}{b} e^{ \frac{b}{m} t} + c \\ v = -g \frac{m}{b} + ce^{- \frac{b}{m} t}$

At terminal velocity: -bv = mg -> v = -mg/b

$v = v_t + ce^{- \frac{b}{m} t}$

At t = 0:

$v = v_0 \\ c = v_0 - v_t$
$v = v_t + (v_0 - v_t)e^{- \frac{b}{m} t} \\ \frac{v - v_t}{v_0 - v_t} = e^{- \frac{b}{m} t} \\ \\t = - \frac{m}{b} ( ln \frac{v}{v_t} - ln \frac{v_0}{v_t} )$

Solve for t at v = 0.02 * v₀

$v_0 = -6.5 \frac{m}{s} \\ v_t = - m \frac{b}{g}$