Question: A ball of mass m is thrown vertically upward with initial velocity $v_0$. Air resistance is proportional to the square of the velocity. If the terminal velocity of the ball is $v_t$, show that when the ball returns to its original position its velocity satisfies


I began by splitting the whole motion in two: i) Upwards and ii) Downwards...

i) Upwards:

Let y be the upward displacement from the starting point y=0, h be the maximum height of the ball, k is a constant.

Using Newton's second law of motion, F=ma, I got:

${mv}\frac{dv}{dy}=mg-mkv^2 \Rightarrow {v}\frac{dv}{dy}=g-kv^2 \Rightarrow \int dy = -\int \frac{vdv}{g+kv^2}$

Once I insert the bounds for the integrals I get:


ii) Downwards:

Now this is where I've gotten stuck and the concept of terminal velocity has gotten me very confused. Can anyone tell me where to proceed from here (if what I've gotten so far is correct) in order to get $\frac{1}{v_1^2}=\frac{1}{v_0^2}+\frac{1}{v_t^2}$



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  • $\begingroup$ Remember that drag switches direction on the downturn. $\endgroup$ – ja72 Oct 29 '13 at 20:15
  • $\begingroup$ Just a simple mistake. You forgot a sign in the first integral. The argument of the integral should be $v/(kv^2-g)$. And maybe you should check that $h>0$ holds. $\endgroup$ – Ana S. H. Oct 29 '13 at 20:18

Nice problem.

I get on the upstroke

$$ h = \int_{v_0}^0 \frac{v}{-g - k v^2}\,{\rm d} v = \frac{1}{2 k} \ln \left (1+\frac{k v_0^2}{g}\right) $$

and the downstroke

$$ -h =\int_{0}^{v_1} \frac{v}{-g + k v^2}\,{\rm d} v = \frac{1}{2 k} \ln \left (1-\frac{k v_1^2}{g}\right) $$

with the terminal velocity $a(v) = -g + k v_t^2 = 0 \Rightarrow v_t = \sqrt{ \frac{g}{k}} $

equating the $h$ on the upside and downside (and setting $k=\frac{g}{v_t^2}$) yields

$$ \frac{v_t^2}{v_0^2+v_t^2} = \frac{ v_t^2-v_1^2}{v_t^2} \Rightarrow \\ \frac{1}{v_1^2} = \frac{1}{v_0^2} + \frac{1}{v_t^2} $$

  • $\begingroup$ Can you explain this line in more detail for me please: with the terminal velocity "$a(v) = -g + k v_t^2 = 0 \Rightarrow v_t = \sqrt{ \frac{g}{k}} $" Thanks $\endgroup$ – Vladimir Nabokov Oct 29 '13 at 22:02

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