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$$\begin{array}{cc} v(t)=v_0+gt&\text{velocity}\\ h(t)=v_0t+\frac{1}{2}gt^2&\text{freefall equation of motion}\\ \Delta t_f=\sqrt{\frac{2h}{g}}&\text{time to free-fall}\\ \Delta t_t=\frac{h}{v_t}&\text{time to fall at terminal velocity}\\ v_t=\sqrt{\frac{2mg}{\rho AC_d}}\stackrel{\text{human}}\approx 56\text{ m s}^{-1}&\text{terminal velocity}\\ t_t=\frac{v_t}{g}\stackrel{\text{human}}=5.7\text{ s}&\text{time to reach terminal velocity} \end{array}$$