A skydiver falls with the quadratic drag dv/dt = g - g/vt²v², where g = 9.8 m/s² and the terminal…

A skydiver falls with the quadratic drag $\dfrac{dv}{dt} = g - \dfrac{g}{v_t^2}v^2$, where $g = 9.8$ m/s$^2$ and the terminal speed is $v_t = 50$ m/s. Use solve_ivp with an event to find the time when the speed first reaches 99 percent of the terminal speed. The exact solution is $v = v_t \tanh(gt/v_t)$, which gives the time $\dfrac{v_t}{g}\,\text{artanh}(0.99)$.