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I've been trying to solve a problem recently:

An object with mass $m$ is thrown at the angle $\alpha$ from the ground with initial velocity $v_0$. At the point of impact vertical projection of velocity turned out to be two times less than initial vertical projection. The resistive force is proportional to velocity. Find total time of flight $T$.

We have a system of differential equations here:

$$\begin{cases}\frac{\text{d}v_x}{\text{d}t} = -\frac{k}{m}v_x\\\frac{\text{d}v_y}{\text{d}t}=g-\frac{k}{m}v_y\end{cases}$$

Given the initial conditions $v_x(0)=v_{0x}$ and $v_y(0)=v_{0y}$ I can find $T$ in terms of $v_x$ and $v_y$. If I equate two these expressions, I'll get an equation with 3 unknown variables $v_x$, $v_y$ and $k$. Also, I know that $v_y(T) = \frac{1}{2}v_{0y}$. This gives me another substitution I can make. But still I'm left with one equation and two unknown variables: $v_x$ and $k$. What other condition should I use to solve this equation for $T$?

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closed as off-topic by sammy gerbil, Chris, Kyle Kanos, Cosmas Zachos, Jon Custer Feb 19 '18 at 16:10

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – sammy gerbil, Chris, Kyle Kanos, Cosmas Zachos, Jon Custer
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  • $\begingroup$ Welcome to Physics! Please note that this site is not a place to obtain solutions to worked problems. Please see this Meta post on asking homework-like questions and this Meta post for "check my work problems". $\endgroup$ – Kyle Kanos Feb 15 '18 at 11:14
  • $\begingroup$ @KyleKanos What is wrong with my question? The link you gave me contains this extract: A good homework question states the problem clearly, shows an attempt to work through it, and identifies the specific issue that is giving the questioner trouble. Isn't my question formed alike? $\endgroup$ – Alex Danilov Feb 15 '18 at 16:22
  • $\begingroup$ no, your question is, "How do I solve this problem" and that is not a concept. IIRC, the post also says If you need to check your work, ask a classmate or your instructor. $\endgroup$ – Kyle Kanos Feb 15 '18 at 16:26
  • $\begingroup$ @KyleKanos I've tried to solve the problem for about a week. Maybe I made a mistake not attaching multiple sheets of paper with my attempts and I will keep that in mind when asking further questions. Still, I thought that I'd made clear what is my issue (I had one equation and two unknown variables). $\endgroup$ – Alex Danilov Feb 15 '18 at 16:36
  • $\begingroup$ @KyleKanos Also, I should have phrased the question differently. Thanks for pointing that out. $\endgroup$ – Alex Danilov Feb 15 '18 at 16:39
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I worked this out as a bit of a review of air resistance and I hope it's correct, but let me know if anything sounds off.

Since the questions are independent and we're interested in the condition $v_y(T)= \frac{v_{0y}}{2}$, let's ignore $v_x$ and the positions $x, y$ in general. I'll make the change $v_y \to u$ so we don't have subscripts everywhere and the definition $\alpha = \frac{k}{m}$. The equation for $u$ is

$$ m \dot{u} = -\alpha u - g $$

We can solve this by a change of variables $w = -\alpha u - g$, so $\dot{w} = -\alpha \dot{u}$ and we have the equation

$$ \begin{align} \dot{w} = -\alpha w &\implies w = w_0e^{-\alpha t} \\ &\implies u = \frac{w_0}{\alpha} e^{-\alpha t} - \frac{g}{\alpha} \end{align} $$

We can solve for $w_0$ in terms of $u_0$ and get $w_0 = -\alpha u_0 - g$, yielding a solution for $u$:

$$ u = \alpha^{-1} \left[ \left( \alpha u_0 + g \right) e^{-\alpha t} - g \right] $$

We then impose our restriction $u(T) = \frac{u_0}{2}$:

$$ \begin{align} &\frac{u_0}{2} = \alpha^{-1} \left[ \left( \alpha u_0 + g \right) e^{-\alpha T} - g \right] \\ &\frac{\alpha u_0 + 2 g}{2 \left( \alpha u_0 - g \right)} = e^{-\alpha T} \\ & T = -\alpha^{-1} \ln{\left[ \frac{\alpha u_0 + 2 g}{2 \left(\alpha u_0 + g \right)} \right]} \end{align} $$

Now we would hope that $\lim\limits_{k \to 0}T = \frac{2u_0}{g}$ to match the case when the resisting force goes away. Unfortunately, if try plugging this into your choice of calculation program, you'll get $\lim\limits_{k \to 0}T = \frac{u_0}{2g}$, so this answer is missing a factor of 4 somewhere (or probably a factor of two twice). I've worked it out a couple of times and can't seem to find where those factors have gone.

It does look like this solution is covered in Taylor's mechanics book (2005) section 2.2 (starting on pg 49 for my version), but the limit on that solution doesn't seem to be working out for me either.

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  • $\begingroup$ Thank you for your answer! The problem is, coefficient ”k” is unknown (at least that’s what people said when they gave me this problem) $\endgroup$ – Alex Danilov Feb 14 '18 at 10:36
  • $\begingroup$ Ahh I seem to have missed that part. I was thinking that maybe an easier way to solve this problem would be using the characteristic time of the exponential to calculate the half-life of the velocity. But that's still going to have a $k$ somewhere I believe. $\endgroup$ – danielunderwood Feb 15 '18 at 19:58

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