Answer
t=15.68 s
Work Step by Step
Equation 1.10:
$v(t)=\frac{gm}{c}(1-e^{-\frac{c}{m}t})$
1) Find the velocity the first parachutist reached in 10 s.
$v_1(10)=\frac{(9.81)(70)}{12}(1-e^{-\frac{12}{70}(10)})\approx46.919217437 \frac{m}{s}$
2) Set the number found in step (1) equal to $v_2(t)$ to find the time it takes for the second parachutist to reach that velocity.
$v_2(t)=\frac{(9.81)(75)}{15}(1-e^{-\frac{15}{75}t})=46.919217437\frac{m}{s}$
3) Solve for $t$.
$\frac{(46.919217437)(15)}{(9.81)(75)}=1-e^{-\frac{15}{75}t}$
$e^{-\frac{15}{75}t}=1-\frac{(46.919217437)(15)}{(9.81)(75)}$
$e^{-\frac{1}{5}t}\approx0.0434410308$
Put a natural log (ln) on both sides:
$ln(e^{-\frac{1}{5}t})= ln(0.0434410308)$
Log properties:
$-\frac{1}{5}tln(e)=ln(0.0434410308)$, and $ln(e)=1$
Therefore:
$t\approx15.68s$
