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I stumbled with this problem in a notebook that has been bothering for the whole day(actually 3)...The answer is written but there's no explanation nor a steb-by-step procedure or anything. If you know how to analyse the problem and create an equation from this, I will be very grateful!

A body with mass m falls somewhere where there's a proportional resistance of |V|r, r is a positive constant. Find the velocity's limit of the body. Answer: V=(mg/k)(1/r) Thank you very much for your time, help and expertise.

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You need to model with a differential equation this physical situation.

Via Newton's second law we know: $$F=ma\ \ \ \ (1)$$ where $F$ is net force, $m$ is the mass of the body, and $a$ is the body's acceleration. But $$a=\frac{dv}{dt},\ \ \ \ \ \ (2)$$ $$F=mq-vr.\ \ \ \ (3)$$ From (1)-(3) we have the next differential equation $$m\frac{dv}{dt}=mg-vr,$$ solving this differential equation we get $$v(t)=\frac{mg-e^{-\frac{r}{m}t}}{r}.$$ Taking limit $t\to \infty$ we get $$v(t)=\frac{mg}{r}.$$

Regards

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From Newton's Law, where the force (in the positive, downward, direction) is due to gravity and the resistive force, we have:

$F = m a = m g - r v$ or $m {\partial^2 x \over \partial t^2} = m g - r {\partial x \over \partial t}$.

The solution of this differential equation is:

$x(t) = g t/r - C_1 e^{-r t}/r + C_2$.

Taking the derivative to get velocity and sett $t -> \infty$ gives your answer.

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HINT The terminal velocity will be achieved when the sum of the forces is zero. The resistance force is $krV$ and there is a force due to earth's gravity: $mg$

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