A heavy particle is projected at speed $U$ at an angle $\alpha$ to the horizontal. The particle is subject to air resistance which is experimentally found to vary proportionally to the square of the speed. Show that

$$\vec{\dot{v}} = -\frac{g}{V^2}\lvert\vec{v}\rvert\vec{v} - g\vec{j},$$

where $V$ is the terminal velocity of the particle. If $\alpha = \frac{\pi}{2}$ (so that the particle is projected directly upwards), find the maxium height reached and the time taken to reach it. What is the speed of the particle when it returns to the horizontal?

I'm having troubles solving this question, firstly I have some fundamental mis-understandings,

  • if the particle is subject to air resistance which is proportional to the square of the speed then do we model newton second law as:

    $$\vec{\ddot r} = \vec{g} - k|\vec{v}|^2\dfrac{\vec{v}}{|\vec{v}|}$$

    or $$\vec{\ddot r} = \vec{g} - k|\vec{v}|^2$$?

    I'm assuming the former by looking as what's required but why? The question says it's just proportional to speed, why in the direction of the velocity?

  • Secondly, why do we have to artificially put a negative sign? Does this mean that $k>0$ from now on?

As for my attempt of the question I done it as follows:

$$\vec{\ddot r} = \vec{g} - k|\vec{v}|^2\dfrac{\vec{v}}{|\vec{v}|} = -g\vec{j} - k|\vec{v}|\vec{v} $$

so we have, $\vec{\dot v} = -g\vec{j} - k|\vec{v}|\vec{v}$

solving this I get the solution of $$ \vec{v} = \dfrac{-g}{k|\vec{v}|} + \vec{c}e^{-kt|\vec{v}|} $$ where $\vec{c}$ is a constant vector,

now from here I get terminal velocity as $\dfrac{-g}{k|\vec{v}|} \vec{j}$ but I am unsure how to get $V^2$ from here.

on second thoughts if $|\underline{v}| = V$ then I get the required result, but why would the speed be the same as the terminal speed


2 Answers 2


You can find out why by asking about when does the falling object reach terminal velocity? We don't know exactly without experimentally doing it, but we do know that if we continue to fall for any amount of time past our terminal velocity, our velocity measured at any point during the aforementioned time will be the terminal velocity. Luckily for you, you already have the function for velocity with drag ( assuming proportionality to it looks like $ v^{2} $. We can turn the idea I put forth into mathematics in the form $$ \lim_{t\rightarrow \infty }\vec{v}(t) $$ What happens to $v$ as $t$ gets large? $ce^{-ktv}$ goes to zero. The only part of your velocity function left is at large $ t $ is $ \frac{-g} {k\vec{v}} $. At large $ t $ your function is equivalently $$ \vec{v}_{term} = \frac{-g} {k\vec{v}_{term}} $$

Solve for $ \vec{v} $, and you have $$ \vec{v}_{term} = \sqrt{\frac{-g}{k}} $$ Where of course $ k $ is your drag constant which has to be experimentally measured, but encodes all the physical characteristics of the object in motion.

This should correspond also to the asymptote of the function when you graph it. That is also your why.


The unit vector and the negative sign are in the "first" equation because you need some way of specifying that the direction of the drag force opposes velocity. Indeed, we must then have $k>0$ Note that the "second" equation can't be correct because it has you adding a vector $\vec{g}$ to a real number $|\vec{v}|^2$.

Look at your expression for terminal velocity. It depends on velocity. Does that strike you as strange?

  • $\begingroup$ Yes I thought that terminal velocity only depends on gravity and the opposing force, why does it depend on the speed the particle? I guess I made a mistake, I'll take another look. $\endgroup$
    – John
    Mar 15, 2014 at 21:29
  • $\begingroup$ on second thoughts if $|\underline{v}| = V$ then I get the required result, but why would the speed be the same as the terminal speed $\endgroup$
    – John
    Mar 15, 2014 at 21:53
  • $\begingroup$ As @DoryanMiller points out, it's the asymptotic value of the speed. You calculated it. Another way to get there is to notice that the terminal speed is the speed at which $\vec{v}$ does not change. That is, $\dot{\vec{v}} = 0$. Use that in your expression for $\dot{\vec{v}}$, and specialize to the case where the object is falling down. $\endgroup$
    – garyp
    Mar 16, 2014 at 2:56

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