Particle's maximum height in liquid

1. Sep 9, 2006

fullerene

I'm not sure if this is the appropriate subforum, but here it goes:

Suppose you have a particle, with the mass m, at the bottom of jar which is filled with liquid. The particle is shot upwards, parallel to the gravitational force, with the speed v0. The drag force is proportional to the particle's speed. What is the maximum height ymax of the particle?

First, "down"force equals to:
Ft=W+Fd
m*a=m*g+b*v
=> a(v)=g+(b/m)v

v(t)=v0+a*t, at ymax: v(t)=0
=> 0=v0+a*t
...

With cross subtituting, I cannot eliminate the variables v or t from the ymax function. Somehow I think the ymax function should not depend on the v or t variables, but maybe I'm wrong. Can you guys help me?

Thanks!

Last edited: Sep 9, 2006
2. Sep 9, 2006

MathematicalPhysicist

i think it should be:
Ft-W-Fd=ma
where Ft is force that needed to shoot the particle.

3. Sep 9, 2006

fullerene

Hmm... I think that the force to shoot the projectile is irrelevant because the projectile has experienced the accelerating process and the force that "shoot" the projectile is no longer acting on it. I dunno, maybe I'm wrong.

Btw, when we consider that this experiment is done in vacumm, the ymax function is obtainable rather easily, just use y(t)=v0*t+(1/2)*g*(t^2). The value of t can be obtainable using v(t)=v0+g*t, which v(t) is zero (at ymax). Then ymax is just a function of v0 -> ymax(v0)=a*v0+b.

4. Sep 9, 2006

MathematicalPhysicist

i think you can use here energies:
$$K_{h}-K_{v_0}=mgh+bv^2/2$$
and the fact that: h=(v^2-v0^2)/2g (also known as galileo equation).
but im not sure about the rhs of the first equation.

5. Sep 9, 2006

HallsofIvy

Staff Emeritus
Since a= dv/dt and depends on v, you can't solve that algebrically. You have a differential equation:
[tex]\frac{dv}{dt}= -g- \frac{bv}{m}[/itex]
That's relatively simple "non-homogeneous linear differential equation with constant coefficients".

6. Sep 9, 2006

fullerene

Ok, thanks HallsofIvy and loop_quantum_gravity.