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An object falls from a height $h$ above water through air with negligible drag. In the water, the upward buoyancy exactly balances the downward gravitation force. The only remaining force on the body in the water is a drag force with magnitude $cv^2$ per unit mass, $c$ is a constant, and $v$ is the velocity. Show that at depth $d \geq 0$ the velocity is $\sqrt{2gh} e^{-cd}$

I'm taking everything downwards to be positive, so above the water I have the equation:

$$v\dfrac{dv}{dy} = g$$

and thus $\int_0^{v_0} vdv = \int_h^0gdy $ where $v_0$ is the velocity when it hits the water. From here I already go wrong, because solving this I get $v_0^2 = -2gh$ which implies the particle goes back up which is absurd. I looked at the solutions and then took everything upwards to be positive instead, and ended up with the integral $\int_0^{v_0} v dv = -\int_h^0 gdy$ which goes on to give the correct answer. Could someone explain how I modelled the particle above the water incorrectly?

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    $\begingroup$ Have you tried setting up a coordinate system? That should help... $\endgroup$
    – dingo_d
    Jan 10, 2014 at 21:04

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First of all $v^2_0=-2gh$ would not imply that the particle moves upward, but that this problem as no real solution $v_0$ or that you made a mistake, which is:

If you set up your coordinate system to point downward, the object in this coordinate system does not move from $h>0$ to $0$ as this would represent a upward motion. It should rather move from $-h<0$ to $0$ or from $0$ to $h$. Thus in both coordinate systems you end up with $$v^2_0=2gh.$$

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