# Projectile motion with drag force of the form $\vec f =-k \vec v$ [closed]

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?

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

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" – BMS, DavePhD, Brandon Enright, Danu, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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. – Floris Oct 2 '14 at 19:16
• @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... – E Be Oct 2 '14 at 20:19
• It's tricky, but wolfram alpha says it's solvable – Floris Oct 2 '14 at 20:45
• 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. – Floris Oct 2 '14 at 21:14
• I suggest that you edit the question to include the work you have done - this is becoming more interesting. – Floris Oct 2 '14 at 21:14

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 $v_f$