## Numerical Methods for Engineers

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: $t\approx15.68s$