4
$\begingroup$

If an object is launched directly up at its terminal velocity, will it have enough time/energy to reach its terminal velocity again before it hits the ground, or will drag prevent this? In this case, the object is an average human, but I'm not sure the specifics matter.

$\endgroup$
  • $\begingroup$ (This question was prompted by a very weird encounter in a roleplaying game, involving teleporting after jumping off very tall buildings.) $\endgroup$ – jpfx1342 May 29 '16 at 1:04
  • $\begingroup$ The answer is no. $\endgroup$ – David Hammen May 29 '16 at 1:36
  • $\begingroup$ An object takes an infinity time to reach terminal velocity. An object returning to the ground after being launched vertically doesn't have an infinite amount of time before hitting the ground, so it can never achieve terminal velocity again. The launch velocity doesn't even matter here. $\endgroup$ – Gert May 29 '16 at 2:36
  • $\begingroup$ its terminal velocity, please? $\endgroup$ – Federico May 29 '16 at 13:12
  • $\begingroup$ My first intuitive answer was that it was no, but I wanted some more details. In any case, we decided the fall/launch would still definitely be fatal. :) $\endgroup$ – jpfx1342 May 30 '16 at 7:16
6
$\begingroup$

The equation of motion considering linear drag is (axis oriented upwards) $$m\frac{dv}{dt}=-mg-bv.$$ The terminal velocity $v_t$ is obtained when $\frac{dv}{dt}=0$, i.e. $$v_t=-\frac{mg}{b}.$$ Solving with the initial condition $v(0)=v_0$ we get $$v(t)=v_t(1-e^{-bt/m})+v_0e^{-bt/m}.$$ If the initial velocity equals (in magnitude) the terminal velocity then $$v(t)=v_t(1-2e^{-bt/m}).$$ Then $v(t)=v_t$ implies $0=e^{-bt/m}$. This equality being met only for $t\rightarrow\infty$, i.e. the particle will not reach its terminal velocity during the fall.

$\endgroup$
  • 1
    $\begingroup$ This true also for a square-of-velocity drag force model. As long as $F_{drag}=f(v)$ it holds. +10. $\endgroup$ – Gert May 29 '16 at 2:41
  • $\begingroup$ @Gert Exactly. I just didn't want to mention quadratic drag because solving the equation of motion is not as simple as in the linear case. $\endgroup$ – Diracology May 29 '16 at 2:46
  • $\begingroup$ You did right. :-) $\endgroup$ – Gert May 29 '16 at 2:47
  • $\begingroup$ I'm accepting this answer, but I should note the other answers are excellent as well. :) $\endgroup$ – jpfx1342 May 30 '16 at 7:12
6
$\begingroup$

Diracology's answer is perfectly right, but I think perhaps another way to answer the question is to explicitly calculate the velocity of the falling particle as it falls back to the altitude it was launched from. Diracology's last equation assumes that the particle can keep falling below the level it was launched from forever and the increase in speed is monotonic, thus we see that the particle's speed when it falls back through its initial altitude is some nonzero amount less than the launch speed.

Method 1 (qualitative calculation): Body's initial kinetic energy is $E_0=\frac{1}{2}\,m\,v_0^2>0$. Gravity force field is conservative. Therefore change $\Delta U$ in gravitational potential by the time the body falls back to its initial altitude is nought. The drag force acts against the body's velocity at all times. Therefore the body does work on the air, so that by the time it reaches its initial height, it must have lost some nonzero amount $E_d$ of its total energy. Thus, measuring potentials relative to the initial altitude, we have $E_0 - E_d + \Delta U = \frac{1}{2}m\,v_f^2$, where $v_f$ is the sought velocity. Therefore we see that $v_f < v_0$ and that the difference is nonzero.

This qualitative method has the advantage of being independent of the functional form $f(v,\,\cdots)$ of the drag (e.g. ram pressure drag varies like $v^2$ and possibly the drag might depend on the position too); the only thing we need to know is that the drag opposes the motion at all times.

Method 2 (quantitative): Write Diracology's equation of motion in the form:

$$m\,v\,\,\frac{\mathrm{d}v}{\mathrm{d} y} = -m\,g - b\,v$$

where $y$ is the altitude. Thus conclude

$$\frac{m}{b}(v-v_0) - \frac{m^2\,g}{b^2} \,\log\left(\frac{m\,g+b\,v}{m\,g+b\,v_0}\right) = -y$$

and find the second solution $v_f$ for $y=0$, i.e. for when the body comes back to the beginning altitude $y=0$ (the first is, of course $v=v_0$). This is a transcendental equation in $v$, but it can be manipulated to show that $v<v_0$ (write the condition for $y=0$ in the form $b\,(v-v_0) = m\,g\log\left(\frac{m\,g+b\,v}{m\,g+b\,v_0}\right)$ and show that the two sides can only have the same sign if either the are nought ($v=v_0$, our first solution) or strictly $v<v_0$ ).

$\endgroup$
  • 1
    $\begingroup$ Fun fact: regardless of the values for gravity, mass, and drag coefficient, the impact velocity as a fraction of the terminal velocity is $1+W(-2e^{-2})\approx 59.4\%$ for linear drag ($W$ is the Lambert W function), and $1/\sqrt{2}\approx 70.7\%$ for quadratic drag. $\endgroup$ – 2012rcampion May 29 '16 at 20:40
5
$\begingroup$

Obviously no, by non-increase of energy (drag always produces negative work). If at some positive height the object would reach its launch velocity (whatever that was, it being the terminal velocity is irrelevant for this argument), it would have obtained its initial kinetic energy and gained potential (gravitational) energy, so some positive work has to be performed. But drag (alone) cannot do that.

(This is basically also said in the answer by Rod Vance, but I feel that if an explanation without formulas exists, it is worth giving.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.