The Drag Force and Terminal Speed — Question 7

Question 7

Opening a parachute changes acceleration immediately, not velocity

Take $g=10$ m/s$^2$ and neglect buoyancy unless explicitly specified. An 80 kg system descends at its old terminal speed of 40 m/s. Its quadratic drag coefficient suddenly becomes four times the old value. Assume mass stays fixed, deployment takes negligible time, and the new law applies immediately without an impulsive force.

1. Infer the old and new drag coefficients.

2. Find the new terminal speed and the acceleration just after deployment.

3. Explain what happens to velocity at deployment and whether the system initially moves upward.

Question 7 - Solution

Opening a parachute changes acceleration immediately, not velocity

(a) Calibrate from the old balance.

Calibrate the old drag law at terminal descent: $$\boxed{c_1=\frac{mg}{v_{t1}^2}=\frac{80(10)}{40^2}=.50\ \mathrm{kg/m}},\qquad \boxed{c_2=4c_1=2.0\ \mathrm{kg/m}}.$$

(b) Distinguish immediate and final states.

The new long-time balance is $$\boxed{v_{t2}=\sqrt{\frac{800}{2}}=20\ \mathrm{m/s}}.$$ Immediately after deployment the speed is still 40 m/s: $$\boxed{a(0^+)=\frac{mg-c_2v(0^+)^2}{m}=\frac{800-2(40)^2}{80}=-30\ \mathrm{m/s^2}}.$$

(c) Keep velocity continuous.

No impulsive force means $$v(0^+)-v(0^-)=\lim_{\epsilon\to0}\frac1m\int_{-\epsilon}^{+\epsilon}F_{\mathrm{net}}\,dt=0.$$ Thus the body initially continues downward at 40 m/s while accelerating upward. For $v>20$ m/s, $\dot v=10-v^2/40<0$; the smooth model approaches 20 m/s from above without an instantaneous reversal.