PARACHUTE PROBLEM 3

(a) 0 ≤ t ≤ 250 seconds  (b) 59 ≤ t ≤ 65 seconds

Fig. 2.1. Motion for (a) 0 ≤ t ≤ 250 seconds and (b) 59 ≤ t ≤ 65 seconds for a piecewise constant coefficient of air resistance.

and x(t_{d}^{+}) = x(t_{d}^{−}) where x(t_{d}^{±}) denote the right- and left-hand limits of the position at t_{d}, respectively. Even though the calculation is rather tedious, the explicit solution is not difficult to obtain;^{2} for t > t_{d} the velocity is

v(t) = \frac{mg}{k_{1}} (e^{− \frac{k_{1}}{m} t_{d}} − 1) e^{− \frac{k_{2}}{m} (t−t_{d})} + \frac{mg}{k_{2}} (e^{− \frac{k_{2}}{m} (t−t_{d})} − 1) . (2.4)

For the sake of this discussion, consider the parameters found in Example 3, p. 112, of Nagle and Saff [7]. Figure 2.1 shows graphical solutions for the position, velocity, and acceleration for 250 seconds after the jump begins as well as a closer look at the motion near the time the chute is deployed.

The graphical solution provides approximate answers to many of the simple questions about the jump. For example, the parachute is opened at an altitude of ≈ 1300m when the velocity is ≈ 49 m/s. Landing occurs a little more than three minutes later, with a velocity of ≈ 7 m/s. More precise answers can be found using the explicit solution (see [5]). Is an impact velocity of ≈ 7 m/s safely survivable without injury?

The degree to which the position appears to be piecewise linear is noteworthy, particularly when compared to the complicated form of the exact solution, (2.2)_{2} and the integral of (2.4). This is easily explained by the rapid convergence of the velocity to the terminal velocity during each stage of the jump. The graph of the velocity immediately following deployment of the chute is so steep that the curious reader might question whether the curve is continuous. (This is a good question for the students.) But, what does this say about the acceleration?

Notice that computing the acceleration by differentiation, a = v′, is complicated by the piecewise definition of the velocity. (What is v′(t_{d})?) A simpler method of obtaining the acceleration is to refer directly to the ODE that governs the velocity (2.1)_{1}. That is, a = −g − \frac{k}{m}v. This approach to the acceleration clearly indicates that there is a jump in the acceleration at the instant the parachute is deployed. The snatch force is the acceleration at the first instant when the assembly reaches full extension; the opening shock, or jerk, is the shock produced while the parachute deploys [8].

2This is a perfect situation to use a computer algebra system. See [5] for a discussion of the use of Maple for the parachute problem.