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

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The discussion focuses on deriving the terminal speed of a falling spherical object using the equation m(dv/dt) = c2v^2 + c1v - mg. It suggests that solving for v as a function of time may be unnecessary since terminal speed is a constant, implying that setting dv/dt to zero simplifies the problem. The conversation emphasizes that the expressions derived are closely related to the provided solution. A recommendation is made to follow a specific approach that clarifies the relationship between the derived expression and the actual solution. Ultimately, understanding terminal speed simplifies the analysis of the falling object's motion.
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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