Show that the terminal speed of a falling spherical object ....

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SUMMARY

The discussion focuses on deriving the terminal speed of a falling spherical object using the equation ##m\frac{dv}{dt} = c_{2}v^2 + c_{1}v - mg##. The contributor suggests that instead of solving for ##v(t)## through partial fractions, one can directly analyze the condition where ##{dv\over dt}=0## to find the terminal speed, which is a constant. The method proposed by user @BvU is highlighted as an effective approach to simplifying the problem and arriving at the correct solution.

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Celso
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Homework Statement
Show that the terminal speed of a falling spherical object is given by ##v_{t} = [ (mg/c_{2})+(c_{1}/2c_{2})^2]^{1/2} - (c_{1}/2c_{2})## when both the linear and the quadratic terms in the drag force are taken into account.
Relevant Equations
##m\ddot x = c_{2}v^2 + c_{1}v - mg##
To write ##v## as a function of time, I wrote the equation ##m\frac{dv}{dt} = c_{2}v^2 + c_{1}v - mg \implies \frac{mdv}{c_{2}v^2 + c_{1}v - mg} = dt##
To solve this, I thought about partial fractions, but several factors of ##-c_{1} \pm \sqrt {c_{1}^2 +4c_{2}*mg}## would appear and they don't show up in the resolution of the problem statement
 
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Terminal speed is a constant ##\Rightarrow## no need to solve for ##v(t)##. Just looking at ##{dv\over dt}=0## is sufficient.
 
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Your expression and the given solution are closely related.

Follow the approach suggested by @BvU at #2 and it pops right out, both your expression and the actual solution.
 

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