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An object is fired with initial velocity $v_0$ at inclination angle $\theta$ above the horizontal. Drag force in the air is taken into account and is of the form $\vec f =-k \vec v$.

Knowing that the velocities in $x$ and $y$ direction are $$v_x(t) = v_{x_0} e^{-\frac{k}{m}t}$$ and $$v_y(t)=-\frac{mg}{k} + (v_{y_0} + \frac{mg}{k})e^{-\frac{k}{m}t}$$ where $v_{x_0}=v_0 \cos \theta $ and $v_{y_0}=v_0\sin \theta$, I want to find the velocity $v_f$ with which the object ends its motion. How could it be found?

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closed as off-topic by BMS, DavePhD, Brandon Enright, Danu, Kyle Kanos Oct 2 '14 at 21:32

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    $\begingroup$ Flippantly: the motion ends when the velocity is zero. More helpfully: you need to find out how long the flight lasts - which you can do by integrating the velocity formula to get y position. See how far you get with that and update your question if you are still stuck. $\endgroup$ – Floris Oct 2 '14 at 19:16
  • $\begingroup$ @Floris: Integrating in the $\hat{y}$ direction to get the position gives $\vec {r}_y=-\frac{mg}{k}t + \frac{m}{k}(v_{y_0} + \frac{mg}{k})(1 - e^{-\frac{k}{m}t})$ and $$-\frac{mg}{k}t + \frac{m}{k}(v_{y_0} + \frac{mg}{k})(1 - e^{-\frac{k}{m}t})=0$$ is not quite solvable... $\endgroup$ – E Be Oct 2 '14 at 20:19
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    $\begingroup$ It's tricky, but wolfram alpha says it's solvable $\endgroup$ – Floris Oct 2 '14 at 20:45
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    $\begingroup$ If you assume that $\frac{k}{m}t$ is small, you can expand the expression and it can be solved on paper. That might make more sense - because the W function is a beast and you have a transcendental ($\approx\text{hard}$) equation there. $\endgroup$ – Floris Oct 2 '14 at 21:14
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    $\begingroup$ I suggest that you edit the question to include the work you have done - this is becoming more interesting. $\endgroup$ – Floris Oct 2 '14 at 21:14
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Integrate $v_y$ over time and solve for when that equals zero. That time is when it hits ground again and will let you find $v_f$

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