Suppose I throw a ball up in the air (with air friction $b>0$) at time $t=0$ and it lands at time $T_1$. So we have the equation:

$$m\frac{\mathrm{d}v}{\mathrm{d}t} = -bv + mg$$

And suppose I throw a ball up in a vacuum (with the same initial velocity) at time $t=0$ and it lands at time $T_2$. So we have the equation:

$$m\frac{\mathrm{d}v}{\mathrm{d}t} = mg$$

Is $T_1<T_2$?

  • $\begingroup$ Try physics.stackexchange.com $\endgroup$ – copper.hat Mar 11 '14 at 19:28
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    $\begingroup$ We wouldn't take this question in its current form on Physics though. Craig, if you want to ask this there, make sure to show what research you've done to try to figure this out yourself, and identify the specific concept that you're stuck on. $\endgroup$ – David Z Mar 11 '14 at 19:37
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    $\begingroup$ It's really a mathematics question. I gave the equations. $\endgroup$ – Craig Feinstein Mar 11 '14 at 20:14

With height as positive in the upward direction, the ODE becomes $\dot{v} = f(v,b) = -{b \over m} v -g$.

Let $t \mapsto v(t,b)$ denote the solution to the ODE for a given $b \ge 0$, and let $t \mapsto x(t,b)$ represent the corresponding height obtained by integrating $v$. Note that the solutions are $C^1$. It is straightforward to write down an explicit solution for both $b=0$ and $b>0$.

The initial conditions are $v(0,b) = v_0 >0$ and $x(0,b) = 0$. We have $v(t,0) = v_0 -gt$, and for $b>0$, $v(t,b) = e^{-{b \over m} t}(v_0 +{mg \over b}) - {mg \over b}$. Integrating gives $x(t,0) = v_0t-{1\over 2} g t^2$, and for $b>0$, $x(t,b) = {m \over b} \left( (1-e^{-{b \over m} t})(v_0 +{mg \over b}) - gt \right)$.

It is straightforward to see that the height is strictly increasing up to some maximum height, and then strictly decreasing after that. In particular, there exist a unique zero crossing for some $t>0$, and that the velocity is strictly negative at this time. Let $T(b)>0$ denote the unique crossing time.

Direct computation gives $T(0) = {2 v_0 \over g}$. We would like to show that $T(b) < T(0)$ for $b >0$.

Suppose $b>0$, then $x(T(b),b) = 0$ gives $ (1-e^{-{b \over m} T(b)})(v_0 +{mg \over b}) = g T(b)$ (with $T(b)>0$, of course). We can rewrite this as $ { v_0 b \over mg } + 1 = { {b \over m} T(b) \over 1-e^{-{b \over m} T(b)} }$.

We will establish the estimate ${x \over 1 - e^{-x}} > 1+{x \over 2}$ holds for $x > 0$. First note that this gives $ { v_0 b \over mg } + 1 > 1+{1 \over 2}{b \over m} T(b)$, from which we obtain $T(b) < {2 v_0 \over g} = T(0)$.

The estimate can be rearranged to ${x-2 \over x+2} > - e^{-x}$, which is clearly true for $x \ge 2$. Now suppose $0<x<2$ and rewrite the inequality as ${2+x \over 2-x} > e^x$. The Taylor series of the left hand side is $-1+\sum_{k=0}^\infty 2 {x^k \over 2^k}$ and the right hand side is $\sum_{k=0}^\infty {x^k \over k!}$. The coefficients of the constant, $x$ and $x^2$ terms are equal, and the left hand coefficient of $x^k$ is ${2 \over 2^k}$ and the right hand coefficient is ${1 \over k!}$ for $k >2$, and since ${2 \over 2^k}>{1 \over k!}$ we see that the inequality holds for $0<x<2$.

  • $\begingroup$ @23rd: Thanks for catching, I have an irreparable sign error which invalidates my answer. $\endgroup$ – copper.hat Mar 18 '14 at 22:39
  • $\begingroup$ Revamped answer. $\endgroup$ – copper.hat Mar 19 '14 at 6:49
  • $\begingroup$ Good job @copper.hat. I actually had thought the answer would be dependent on $m,g$, and $b$. $\endgroup$ – Craig Feinstein Mar 19 '14 at 16:00
  • $\begingroup$ @CraigFeinstein: Well, it is dependent on $m,g$, but here we are comparing the results of $b=0$ with $b>0$ for the same $m,g$. What I mean is that we should write $v(t,b,m,g)$, etc, but that would add a lot of clutter. And we are only interested in whether $T(b,m,g,...) < T(0,m,g,...)$. I had hoped to prove that $T(b)$ (as above) is a strictly decreasing function of $b$ using the implicit function theorem, but got lost in estimates. The problem was trickier than I originally thought. $\endgroup$ – copper.hat Mar 19 '14 at 16:16
  • $\begingroup$ I meant that I had thought that whether $T_1<T_2$ depends on $b,m,g$. $\endgroup$ – Craig Feinstein Mar 19 '14 at 19:03

I remember the answer to this because I was a bit surprised when I first did it.

You'll find that the time spent in the air is less than if it were in a vacuum. Let $v_0$ be the initial velocity and $-v_1$ be the final. You can show without so much work that the time taken for a ball to travel up and back down in a vacuum is given by $$\tau=\frac{2v_0}{g}.$$ When the ball travels up, we have $-mg-bv=F$ and we get $$\int^{t_1}_0dt=-\int^0_{v_0}\frac{v}{g+\frac{b}{m}v}dv=\frac{m}{b}\int^{v_0}_0\left(1-\frac{g}{1+\frac{b}{m}v}\right)dv,$$ which means $$t_1=\frac{m}{b}\log\left(1+\frac{bv_0}{mg}\right).$$ It's the same for travelling back down again, and the time $t_2$ required to do this is $$t_2=-\frac{m}{b}\log\left(1-\frac{bv_1}{mg}\right).$$ You get $$t_1+t_2=\frac{m}{b}\log\left(\frac{mg+bv_0}{mg-bv_1}\right)=\frac{v_0+v_1}{g}.$$ Since $v_1$ must be less than $v_0$ you find that $t_1+t_2<\tau$ which means that the answer to your question is yes.

As commented above however, often you have an equation dependent on $v^2.$ In this case, you would find the speed to be given by a formula like $$v(t)=\sqrt{\frac{mg}{b}}\frac{c_1e^{2\sqrt{\frac{bg}{m}t}}-1}{c_1e^{2\sqrt{\frac{bg}{m}t}}+1}=\sqrt{\frac{mg}{b}}\tanh\left(\sqrt{\frac{bg}{m}}t+c_2\right),$$ as opposed to forces of the form $F_0-bv,$ which give formulae like $$v(t)=v_0e^{-\frac{b}{m}t}+\frac{F_0}{b}\left(1-e^{-\frac{b}{m}t}\right).$$

  • $\begingroup$ On a related note, something you might find interesting is that if you were to throw your ball at an angle with a drag force like you mentioned above (ie. of the form $-bv$), and its horizontal distance travelled is $x$ and vertical $y,$ to maximise $x$ when $y$ is at its peak you would throw it at an angle $$\frac{\sqrt{5}-1}{2}=\frac{1}{\varphi}.$$ $\endgroup$ – Hobbyist Mar 15 '14 at 6:31
  • $\begingroup$ I realise that it is true, but you haven't shown that $|v_1| < |v_0|$. $\endgroup$ – copper.hat Mar 15 '14 at 7:07
  • $\begingroup$ I didn't know if I should have or not. On the one hand, mathematicians are the main people here, so they'll want everything to be a bit more thorough. On the other, I don't like spoon-feeding answers and I thought that it would be clear to OP if OP had studied physics, so I wouldn't want to patronise by showing too much. I'll definitely add it if it's not clear to OP though. $\endgroup$ – Hobbyist Mar 15 '14 at 7:14
  • $\begingroup$ I don't follow the part where you say, "When the ball travels up, we have..." $\endgroup$ – Craig Feinstein Mar 18 '14 at 17:01
  • $\begingroup$ I was unable to show $|v_1|<v_0$ without showing $T_1<T_2$ first :-(. $\endgroup$ – copper.hat Mar 19 '14 at 7:15

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