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q.

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 on the part where "If $\alpha = \frac{\pi}{2}$ ..."

seeing as $\alpha = \frac{\pi}{2}$ it would be acceptable to think of motion in one dimension, so from the "show that" equation I get $$\ddot r = \dot v = \dfrac{-g}{V^2}v^2 - g $$

Now I'm curious here, can I just integrate w.r.t to t? Usually in vector form v is a function of t, and I'm asuming it is here aswell, but is this allowed:

$$\ddot r = \dfrac{-g}{V^2}v^2 - g $$ $$\dot r = \left ( \dfrac{-g}{V^2}v^2 - g \right )t + C$$

If so, why? If not, why not? Furthermore, I get a complex solution to the terminal velocity, what does this mean - As the terminal velocity can't be 0, what can we conclude from that?

edit: solving $\dot v = -kv^2 - g$ where $k = \dfrac{g}{V^2}$ I get

$\dfrac{dv}{kv^2+g} = -dt$ and solving this I get $v = \dfrac{\sqrt{g}}{\sqrt{k}} arctan(C\sqrt{k}\sqrt{g} - \sqrt{k}\sqrt{g}t)$ where C is a constant, integrating this again I get $r = \dfrac{1}{k}\log (sec(c\sqrt{k}\sqrt{g} - \sqrt{k}\sqrt{g}t)) + C_1$ then for $v = 0$ I get $ t = C$ so $r = C_1$, fro r = 0 I get $t = \dfrac{C\sqrt{k}\sqrt{g} - arcsec(e^{-Ck})}{\sqrt{k}\sqrt{g}}$ so $v = \dfrac{\sqrt{g}}{\sqrt{k}}tan(arcsec(e^{-Ck})) = \dfrac{\sqrt{g}}{\sqrt{k}}\sqrt{1 - \frac{1}{e^{-2Ck}}}e^{-Ck}$

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As you said in your question, since $v$ is a function of $t$, you cannot integrate by just multiplying by $t$. The strategy to use here to solve this ODE would be separation of variables. However, you must separately consider the cases of the particle traveling upward and downward, so your second equation is not quite right. If the particle is traveling upward, that is accurate, since the drag force should push it back down. If the particle is traveling down, the drag force should be upward, and therefore positive. Then, the equation would be $$ \dot{v} = g ([v/V]^2 - 1 )$$ The terminal velocity should be immediately obvious. To get the maximum height, just use the upward traveling case and find when $v=0$.

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  • $\begingroup$ are you sure about seperation of variables? I seem to get a pretty horrible equation. My plan was to solve for v, then solve for r. Then for the maximum high to be reached I was planning to let $v = 0$ then find $t$, subbing this back into $r$, but doing this seems much more complicated $\endgroup$ – John Mar 16 '14 at 14:33
  • $\begingroup$ That's the idea. I get $ -g dt = dv/([v/V]^2 + 1)$ $\endgroup$ – ZachMcDargh Mar 16 '14 at 14:39
  • $\begingroup$ I get the same, I have updated my working - it seems too messy and I really feel as though I'm missing something. Could you please explain why the terminal velocity should not be complex? $\endgroup$ – John Mar 16 '14 at 14:55
  • $\begingroup$ I am updating my answer. What I said was not quite complete. $\endgroup$ – ZachMcDargh Mar 16 '14 at 15:57

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