Baseline. Along one axis, m dv/dt = F + m g − b v − c |v| v. With only linear drag, v(t) = v∞ + (v₀ − v∞) e^(−bt/m), v∞ = (F + mg)/b. With only quadratic drag and v₀ = 0, v(t) = v∞ tanh(g t/v∞), v∞ = √(mg/c); the distance is s = (m/c) ln cosh(ωt), ω = c v∞/m.

What this lab adds. Both drag laws can act together (the terminal speed is the positive root of c v² + b v = F + mg), thrust and gravity can be mixed, and a parachute is a quadratic-drag boost c(t) = c + c_p·smoothstep((t − t_d)/t_o) that opens over a chosen time. Its size is in multiples of a reference quadratic coefficient (the case's own c, or the value that matches its linear drag at the linear terminal speed).

Honest simplifications. Constant air density and drag coefficients, one-dimensional motion, no added-mass or buoyancy effects, no opening shock modelling beyond the smooth ramp. Cases given only a quadratic drag factor per unit mass are run with m = 1 kg.

Independent checks. RK4 is compared with the closed forms above for the case's own drag law, with an 8× smaller step for your current settings, and with the root of the terminal-speed equation after the parachute has fully opened. A separate Python test repeats these with its own integrator.

Drag and Terminal Speed Case Studies