F_d = Cd A rho v²/2,
where rho = air density, Cd is a nondim. drag coefficient of order one, and A the cross sectional area of the body. Vertical axis points upward, thus the drag force has positive sign when the object moves down (but only then). Gravitational acceleration is negative and equal -g. When acceleration of drag Fd/M (where M = mass of the body) reaches the same absolute value as gravity's acceleration
g = 9.81 m/s²,
then the body reaches terminal velocity v=V and is falling through air without accelerating or decelerating. So all those detailed properties of the object and air are conveniently contained in one single number, V (terminal velocity). In practice it ranges from ~10 m/s for small bodies that tend to have large A/M ratios, to 100 m/s or more for slender bodies with smaller A/M. You will just assume that in a given problem V is known, and use a nondimensional velocity defined as
u = v/V
Derive the equation of motion using u, g, and V. Show that in both upward and downward motion
du/dt = -(g/V) (1+u|u|).
The free-fall starts at time t=0 with u=0 (initial conditions). Find the exact solution for u(t). Sketch it in a diagram. Use the nondimensional time on the horizontal axis: gt/V. You may use the calculator and compute a few points, to get the shape right.
Is the convergence to the terminal speed at large times exponential or algebraic? What is the time scale? Next, find the vertical distance traveled. When u < 0, 1+u|u| = 1-u². Instead of integrating v(t) = V u(t) over time to find z(t), use the following shortcut method:
dz = v dt, while
dt = -(V/g) du/(1-u²) by separating variables in the equation of motion, so
dz = -(V²/g) u du/(1-u²).
You can integrate the last ODE much more easily than dz = v(t) dt. True, you'll get z(u) not z(t), but u = u(t) is already known, so if you ever need z(t), you can get it as z(t) = z(u(t)). But you don't need z(t) in this problem. You may assume z=0 at u=t=0, for simplicity. Distance z(u) will then be negative. (If needed, the initial height can easily be redefined to be > 0).
Formulate and integrate the equation of motion to obtain u(t), with initial condition u(t0) = u0 > 0, at some negative time t0, and final velocity equal v=0 at time t=0. These final conditions match the initial conditions of the fall (prev. problem). Sketch u(t) or, more precisely, u(gt/V).
Formulate the ODE for the distance traveled, and solve it to obtain z_max = z(u=0)=z(t=0) in terms of V, g, and u0.
Show that the maximum height of a vertical throw in air is always smaller than in vacuum (to obtain the latter, simplify your solution in the appropriate limit of u0, corresponding to very small air density and drag force). For example, how much smaller is z_max in air than in vacuum if u0 = 1 in both cases (upward initial velocity = terminal velocity)?
What comes up must come down and, in our problem, cover the same vertical distance |z| = z_max. Given some u0, what is the final nondimensional speed u, with which the body strikes the ground? Check that the final speed u satisfies |u| < u0, i.e. |v| < v0 (give a physical reason for this) and that during the descent |u| < 1 (why?).
Does the fraction u/u0 depend on anything (u0, g, V?).
Show that u/u0 is the same for all bodies thrown on all planets in the
universe, and only depends on u0= v0/V.