The Drag Force and Terminal Speed — Question 10

Question 10

A terminal-speed fraction requires a finite fall distance

Take $g=10$ m/s$^2$ and neglect buoyancy unless explicitly specified. A body falls from rest under quadratic drag in still fluid. Its terminal speed is 20 m/s. Downward distance y starts at zero. A proposed spatial relation is $v^2=v_t^2[1-e^{-2gy/v_t^2}]$. You may use $d(v^2)/dy=2a$ during downward motion.

1. Verify the relation against $a=g(1-v^2/v_t^2)$ and the initial condition.

2. Find the distance needed to reach 90 percent of terminal speed.

3. Compare with the no-drag distance to the same speed and explain whether exact terminal speed occurs at finite y.

Question 10 - Solution

A terminal-speed fraction requires a finite fall distance

(a) Differentiate with respect to position.

Differentiate the proposed squared-speed relation: $$\frac{d(v^2)}{dy}=2g e^{-2gy/v_t^2}\Longrightarrow a=\tfrac12\frac{d(v^2)}{dy}=g e^{-2gy/v_t^2}=g\left(1-\frac{v^2}{v_t^2}\right).$$ At $y=0$, $v^2=v_t^2(1-1)=0$, satisfying release from rest.

(b) Solve for the remaining squared-speed gap.

The target speed is $v=.9v_t=18$ m/s: $$.81=1-e^{-2gy/v_t^2}\Longrightarrow y=-\frac{v_t^2}{2g}\ln(.19).$$ $$\boxed{y=-20\ln(.19)=33.2146\ \mathrm m}.$$

(c) Compare like target speeds.

Without drag, constant-acceleration kinematics gives $$\boxed{y_{\mathrm{no\ drag}}=\frac{18^2}{2(10)}=16.2\ \mathrm m}<33.2146\ \mathrm m.$$ Drag reduces the acceleration at nonzero speed. Since $e^{-2gy/v_t^2}>0$ for every finite $y$, the proposed law has $v<v_t$ at every finite distance; exact terminal speed occurs only as $y\to\infty$.