For the free-falling bungee jumper with linear drag (Prob. 1.4), assume a first jumper is 70 kg and has a drag coefficient of 12 kg/s. If a second jumper has a drag coefficient of 15 kg/s and a mass of 75 kg, how long will it take her to reach the same velocity jumper 1 reached in 10 s?
Consider
$$ v(t)=\frac{g \cdot m}{c} \cdot\left(1-e^{-\left(\frac{c}{m} \cdot t\right)}\right) v(t)=\frac{g \cdot m}{c^{\prime}} \cdot\left(1-e^{-\left(\frac{\varepsilon^{\prime}}{m} t\right)}\right) \ldots \ldots \text { (1) } $$
\(\mathrm{v}(\mathrm{t})\) - Velocity
g- Acceleration due to gravity \(=9.8 \mathrm{~m} / \mathrm{s}^{2}\)
m- Mass \(=70 \mathrm{~kg}\)
\(c^{\prime}\) - Drag coefficient= \(12 \mathrm{~kg} / \mathrm{s}\)
\(\mathrm{t}-\) Time \(=10 \mathrm{~s}\)
Substitute the values in Equation (1)
$$ \begin{aligned} v(t)=\frac{9.8 \cdot(70)}{12} \cdot\left(1-e^{-\left(\frac{12}{70} 10\right)}\right)^{v(t)} &=\frac{9.8(70)}{12} \cdot\left(1-e^{-\left(\frac{12}{70} \cdot 10\right)}\right) \\ &=46.8714 \mathrm{~m} / \mathrm{s} \end{aligned} $$
Use the value of \(v(10)=46.8714 \mathrm{~m} / \mathrm{s}\) to get the time taken by bungee jumper #2
Substitute in Equation (1)
$$ \begin{aligned} 46.8714 &=\frac{9.8(75)}{15}\left(1-e^{-\frac{15}{75} t}\right) \\ 46.8714 &=49-49 e^{-0.2 t} \\ 49 e^{-0.2 t} &=49-46.8714 \\ 49 e^{-0.2 t} &=2.1286 \\ e^{-0.2 t}=0.043441 \end{aligned} $$
Take \(\ln\) on both sides
$$ \begin{aligned} -0.2 t &=\ln (0.043441) \\ -0.2 t &=-3.136352 \\ t &=\frac{3.136352}{0.2} \\ t &=15.68176 \end{aligned} $$
Therefore, \(t=16 \mathrm{sec}\)