# Particle thrown straight up, need to find velocity

1. Sep 16, 2010

### Mugged

Ok so I have this particle of mass m that is thrown upward with initial velocity $$v_{0}$$. Now there is a downward force due to gravity as well as a square-law drag force described by b*v^2 where b is some constant. I need to find the velocity as a function of time.

Any help is appreciated!

2. Sep 16, 2010

### gabbagabbahey

3. Sep 16, 2010

### Mugged

heres what i have so far:

F = ma = mg + bv^2

a = g + (b*v^2)/(m)

then i have the kinematic equation:
V = V0 +at

so i plug in my acceleration and then i get the following:

V = V0 + (g + (b*v^2)/(m) ) t

but i need to solve for V and im not sure how to algebraically move V to one side? Im stuck

4. Sep 16, 2010

### gabbagabbahey

Do the +/- signs make sense for that? When the particle is moving upwards, the drag force should be downwards and vice versa.

That equation is for constant acceleration problems. Your acceleration depends on velocity (thanks to the drag force) and isn't constant. What is the more general relationship between acceleration and velocity?

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5. Sep 16, 2010

### Mugged

oh ok,

so then i should have a = g - (b*v^2)/(m) with g = -9.8 m/s^2

this is while the particle is moving upwards.

as for a general acceleration to velocity relationship, im not quite sure what you mean. The only other one i can think of is if i differentiated velocity with respect to time i would get acceleration with respect to time.

so since i have acceleration above, you're saying i should integrate to get velocity? if thats correct, i still have no idea how to integrate the right hand side of my above equation

well if i was doing differential equations, it would look something like this:

V' = g - (b*V^2)/(m)

but, im terrible at dif eq, so id need help here too

6. Sep 16, 2010

### gabbagabbahey

If your convention is that v is positive when the particle is moving upwards and negative when it is moving downwards, then yes, that's correct. Similarly, once the particle starts moving downwards, you will have $a=g+\frac{bv^2}{m}$. You can combine these into one expression by using the sign function $\text{sgn}(v)\equiv \frac{v}{|v|}$ :

$$a= g - \text{sgn}(v)\frac{bv^2}{m}$$

Yes, in other words you have a differential equation:

$$\frac{dv}{dt}= g - \text{sgn}(v)\frac{bv^2}{m}$$[/tex]

This ODE is separable, surely you know how to solve one of those?

7. Sep 16, 2010

### Mugged

$$dv= g - \text{sgn}(v)\frac{bv^2}{m} * dt$$

$$v = (g - \text{sgn}(v)\frac{bv^2}{m} ) t$$

is this what you meant? but id still like to get v on one side

8. Sep 16, 2010

### gabbagabbahey

Divide both sides by $$g - \text{sgn}(v)\frac{bv^2}{m}$$ before you integrate.

9. Sep 16, 2010

### Mugged

hey gab, i think i got the answer in the end after i separated, thanks.

by chance, do you have any idea how to find velocity as a function of the displacement?

10. Sep 16, 2010

### gabbagabbahey

Use your expression for v(t) to integrate and get y(t) and then eliminate "t" from the two expressions.

11. Sep 16, 2010

### Mugged

that cant be possible..i have something extraordinarily ugly for my velocity function

$$v(t) = \sqrt{\frac{mg}{b}} tan ( tan^-1( \sqrt{\frac{b}{mg}}V_{0}) - gt \sqrt{\frac{b}{mg}} )$$

oh by the way gab, this function is only valid for when velocity is in the positive upward direction, i realized in my homework that i dont need to account for the velocity going down.

12. Sep 16, 2010

### vela

Staff Emeritus
That only looks complicated. It's actually just a shifted tan function. Let

$$\alpha = \sqrt{\frac{mg}{b}}$$

Note $\alpha$ is just a convenient combination of constants. You can write v(t) as

$$v(t) = \alpha \tan(\phi_0-\frac{gt}{\alpha})$$

where $\phi_0 = \tan^{-1}(v_0/\alpha)$, which is also a constant.