Solving Projectile Motion in Fluid: How Far Will It Go?

[Anders0304]

If you fire a projectile in a fluid, say water. How far will it go before the water resistance stops it.

I know that water resistance is F=½*ρ*v^2*A*C_d
If I know the kinetic energy of the projectile my first impression was that it would simply be the work done by the water resistance vs the energy of the projectile:
E=½mv^2 and W=F*d, where d is distance traveled. But since the velocity of the projectile is reduced with distance because of the water resistance, the water resistance itself is lowered because of the lower velocity. It seems like one variable is dependent on the other. Is there any way to solve this?

also thanks in advance

 

[marawan]

I am currently facing the same issue and am wondering, Have you come up with an answer to your own question?

 

[boneh3ad]

Sure, just set it up as a differential equation assuming you know all the parameters like C_D.

 

[marawan]

How, could you please post some steps?

 

[brainpushups]

The physical set up of the question is unclear, but if the motion is 1 dimensional and depends on inertial fluid drag only then the particle's position is logarithmic with time and the particle will not come to rest in a finite amount of time. You could ask the question as: how far will it go once its velocity is reduced by 99% or something.

 

[marawan]

The physical set up of the question is unclear, but if the motion is 1 dimensional and depends on inertial fluid drag only then the particle's position is logarithmic with time and the particle will not come to rest in a finite amount of time. You could ask the question as: how far will it go once its velocity is reduced by 99% or something.

I have tried all these suggestions and have just had arithmetic problems could u please clarify what u mean with steps? You can assume you know both the mass and shape of the object (rectangular in nature) and since you have both the initial velocity and final velocity it seems like there must be a way to find out the distance the object travels before coming to rest.

thanks in advance

 

[boneh3ad]

...if the motion is 1 dimensional and depends on inertial fluid drag only then the particle's position is logarithmic with time and the particle will not come to rest in a finite amount of time. You could ask the question as: how far will it go once its velocity is reduced by 99% or something.

Of course, this is not true in real life, only in the mathematical model. The dissipative effects of viscosity see to that.

How, could you please post some steps?

Are you familiar with differential equations? The problem with the above approach is that you start with work, which requires your final answer and you end up with two unknowns (##d## and ##v##) and only one equation. Instead, start with a force balance, ##\Sigma F = ma##. Assuming it is one dimensional and the only force is drag, then you know both sides and can set up the equation
[tex]m\dfrac{d^2 x}{dt^2} = \dfrac{1}{2}C_D \rho A v^2 = \dfrac{1}{2}C_D \rho A \left(\dfrac{dx}{dt}\right)^2.[/tex]
So, you've got
[tex]\dfrac{d^2 x}{dt^2} - \dfrac{C_D\rho A}{2m} \left(\dfrac{dx}{dt}\right)^2 = 0.[/tex]
You can solve that with a change of variables,
[tex]v = \dfrac{dx}{dt}.[/tex]
The end result is like @brainpushups said, that the the position is logarithmic with time, and with this "simple" model, it will never truly stop.

 

[Chestermiller]

Of course, this is not true in real life, only in the mathematical model. The dissipative effects of viscosity see to that.
Are you familiar with differential equations? The problem with the above approach is that you start with work, which requires your final answer and you end up with two unknowns (##d## and ##v##) and only one equation. Instead, start with a force balance, ##\Sigma F = ma##. Assuming it is one dimensional and the only force is drag, then you know both sides and can set up the equation
[tex]m\dfrac{d^2 x}{dt^2} = \dfrac{1}{2}C_D \rho A v^2 = \dfrac{1}{2}C_D \rho A \left(\dfrac{dx}{dt}\right)^2.[/tex]
So, you've got
[tex]\dfrac{d^2 x}{dt^2} - \dfrac{C_D\rho A}{2m} \left(\dfrac{dx}{dt}\right)^2 = 0.[/tex]
You can solve that with a change of variables,
[tex]v = \dfrac{dx}{dt}.[/tex]
The end result is like @brainpushups said, that the the position is logarithmic with time, and with this "simple" model, it will never truly stop.

Shouldn't there be a minus sign in your first equation:
[tex]m\dfrac{dv}{dt} = -\dfrac{1}{2}C_D \rho A v^2 [/tex]

Also, this equation can be integrated immediately to get v:
$$v=\frac{v_0}{1+v_0kt}$$
where
$$k=\frac{C_D\rho A}{2m}$$

Chet

 

[boneh3ad]

Shouldn't there be a minus sign in your first equation:
[tex]m\dfrac{dv}{dt} = -\dfrac{1}{2}C_D \rho A v^2 [/tex]

Also, this equation can be integrated immediately to get v:
$$v=\frac{v_0}{1+v_0kt}$$
where
$$k=\frac{C_D\rho A}{2m}$$

Chet

Yes and yes. I really ought to stop doing math at 2 am.