Marion and Thornton example problem, chapter 2.

A projectile subject to a drag force proportional to velocity

m Vx-dot = -k m Vx,            (1)

mVy-dot = - k m Vy - m g,   (2)

Solving (1) gives Vx = Vxo exp(-kt);     x= Vxo/k (1-exp(-kt)).

Solving (2) begins with dVy/( kVy + g) = - dt.

Then k dVy/(kVy + g) = -kdt and if w=kVy + g,   dw/w = -kdt.

This gives kVy+ g = (kVyo + g) exp(-kt). Solving for y(t) gives

y = -g/k t +(1/k^2)(kVyo + g)(1-exp(-kt). Want to solve this for y=0, so

t g/k = (k Vyo + g)(1/k^2) (1-exp(-kt)), or kt = (1+kVyo/g)(1-exp(-kt)).

let T = kt, and s = kVyo/g, both dimensionless parameters.

Then T = (1+s)(1-exp(-T)).

Molasses limit

If s is large (k is large; the 'molasses' limit of very large damping) then T =s, and is large.

kt = kVyo/g, or t=Vyo/g

x= Vxo/k (1-exp(-T)). For larger and larger k, x gets smaller and smaller.

kt = k Vyo/g, or t = Vyo/g, just half of the time it would take without damping.

Rearranging gives 1/(1+s) = (1-exp(-T))/T.

In the 'vacuum' limit, k is small, s is small, and so is T. Expanding each side we get

1-s+s^2+.... = 1 -1/2 T +1/6 T^2 + ... (through 2nd order in smallness).

To first order, we find T=2s, or kt = 2 Vyo k/g -> t = 2 Vyo/g. (the undisturbed flight time)

To second order, we have T = 2(s - s^2 +1/6 T^2).

On the right for T we use the first order value of T, namely T=2s, and substitute this in the righthand side of the expression above to get T = 2(s - s^2+4/6 s^2). Then

T = 2(s+1/3 s^2), and    kt = 2[ Vyo k/g -1/3 (Vyo k/g)^2 ]

finally, t = 2 Vyo/g (1- 1/3 Vyo k/g).

This solution employed a perturbation approach when substituting on the right a first order solution for T in order to be correct to second order in smallness.

The effect of damping should reduce t slightly, since in the molasses limit of large damping t is only half of the 'free-flight' time.