The hailstone of and obeys (downward negative) and has terminal speed

PHYSICS · Subjective · Fermi University Physics · Computational Physics · Mechanics & Dynamical Systems · Kinematics and Newtonian dynamics of particles

Computational Physics, Ch A falling body that knows about air: the first simulation

Problem

The hailstone of m=6.0×10−5 kgm = 6.0\times10^{-5}\,\mathrm{kg} and b=5.9×10−6 kg m−1b = 5.9\times10^{-6}\,\mathrm{kg\,m^{-1}} obeys mv˙=−mg+bv2m\dot v = -mg + bv^2 (downward negative) and has terminal speed v∞=9.9881 m s−1v_\infty = 9.9881\,\mathrm{m\,s^{-1}}. Suppose it is released not from rest but already moving downward faster than this, at v0=−15 m s−1v_0 = -15\,\mathrm{m\,s^{-1}}. Use g=9.81 m s−2g = 9.81\,\mathrm{m\,s^{-2}} and the update rule vn+1=vn+a(vn)Δtv_{n+1} = v_n + a(v_n)\Delta t with a(v)=−g+(b/m)v2a(v) = -g + (b/m)v^2. (a) Hand-compute the velocity after one step of Δt=0.50 s\Delta t = 0.50\,\mathrm{s}. (b) Hand-compute the velocity after two steps of Δt=0.25 s\Delta t = 0.25\,\mathrm{s} each (the same total elapsed time as part (a)), recomputing the drag term at each step. (c) State, from the two numbers, whether the computed speed has already fallen below the terminal speed's magnitude (9.9881 m s−19.9881\,\mathrm{m\,s^{-1}}) at Δt=0.50 s\Delta t = 0.50\,\mathrm{s}, and whether it has at Δt=0.25 s\Delta t = 0.25\,\mathrm{s}.

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