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## Homework Statement

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An object of m-mass is to be thrown from xy-plane with an initial velocity ##\mathbf v_0 = v_0\mathbf e_z \, (v_0 > 0)## to a force field ##\mathbf F = -F_0 e^{-z/h}\mathbf e_z\,## , where ##F_0, h > 0## are constants. By what condition does the object return to ##xy##-plane? How does the situation change if a velocity dependent drag force affects to the object?

## Homework Equations

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##\mathbf F = -F_0 e^{-z/h}\mathbf e_z##

##\mathbf v_0 = v_0\mathbf e_z##

##F =ma = mv\frac{dv}{dz}##

## The Attempt at a Solution

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The sign is positive from xy-plane to upwards as a convention.

There is an acceleration associated with the force field. So Newton's second law should give:

##-F_0 e^{-z/h}\mathbf e_z = m\mathbf a##

##-F_0 e^{-z/h} \mathbf e_z = ma\, \mathbf e_z##

##F_0 e^{-z/h} = -ma##

Integrating ## 0 \to z_0##

[tex]\int_0^{z_0} -F_0 e^{-z/h}\, dz= \int_0^{z_0}m \frac{d^2 z}{dt^2}\, dz[/tex]

[tex]hF_0 \int_0^{z_0} -\frac{1}{h}e^{-z/h}\, dz= m\int_0^{z_0} \frac{d^2 z}{dt^2}\frac{dz}{dt}\, dt[/tex]

[tex]hF_0 \int_0^{z_0} -\frac{1}{h}e^{-z/h}\, dz= m\int_0^{z_0} d \left [\frac{1}{2}\left ( \frac {dz} {dt} \right)^2 \right][/tex]

[tex]hF_0 (e^{-z_0/h}-1)= m\frac{1}{2}(0^2 - \mathbf v{_0}^2)[/tex]

[tex]hF_0 (1-e^{-z_0/h})= m\frac{1}{2}v{_0}^2\mathbf {e_z}^2[/tex]

[tex]v_0 = \sqrt {\frac{2hF_0(1-e^{-z_0/h})}{m}}[/tex]

So I think this velocity should be less or equal to this..

[tex]v_0 \leq \sqrt {\frac{2hF_0(1-e^{-z_0/h})}{m}}[/tex]

My book does not give any answer to this problem, so I'm not 100% sure this is the correct answer but I assume it is.

Then the second part of the question adds a velocity dependent drag force, so I use ##k \mathbf v = - kv\, \mathbf e_z## to represent this force. The Newton's second law leads to:

##-F_0 e^{-z/h}\, \mathbf e_z + k \mathbf v= m \mathbf a##

##-F_0 e^{-z/h}\, \mathbf e_z - kv\, \mathbf e_z= ma\, \mathbf e_z##

##F_0 e^{-z/h} + kv = -mv\frac{dv}{dz}##

First of all I want to know am I even on the right track for trying to solve this

[tex]m\frac{dv}{dz}+\frac{F_0}{v} e^{-z/h} + k = 0[/tex]

Thanks for reading and any help is appreciated

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