# A little bit beyond projectile motion.

1. Oct 10, 2004

### faust9

Ok, here's the deal. I have a projectile motion project that goes a little beyond the normal "ideal projectile" problems.

The project consisted of shooting a projectile from a device that, within a few percent, launched a projectile at the same initial velocity every time it is fired. The launcher has an adjustable launch angle and shoots a ping-pong ball sized 5g projectile with a rough surface (yellow foam ball). My team shot the ball ten times at 15,30,45, and 60 degrees and then measure the distance traveled. We were told this data will be used later this week to find the drag coefficient of the ball but for today I simply need to develop a general equation of motion given:

$$v_0=100m/s,\ \theta=30^\circ,\ m_{proj}=20kg$$ and various values of K.

Also, the force of drag is described as $$F_D=-Kv^2$$

where K is the drag coefficient and v is velocity.

I need to develop the equations of motion in terms of x and y for this "test" projectile which I will use later on to find the drag coefficient of our yellow ball and predict the range of the ball given an angle of launch.

I know I need 2 second order ODE's which I can convert to four first order ODE's, but I don't know where to start or how to start.

I have my FBD--a round ball with weight parallel to the y axis and drag opposite to the direction of velocity.

Any help would be greatly appreciated here. Thanks in advance.

2. Oct 10, 2004

### TenaliRaman

what equations did u get from ur FBD?
Also give ur thoughts on how u think u would like to proceed?

-- AI
P.S -> Man this is fun .... i wish i had done physics this way .... :(

3. Oct 10, 2004

### faust9

That's the thing. We are told to express our equations of motion in terms of x/y but I don't know how to do that. My I get the following from my FBD:

i: $$m\vec a=-\vec F_{Dx}$$

j: $$m\vec a=-\vec F_{Dy} \sin - \vec W$$

I know that $$\vec a=\ddot {x} i+\ddot {y} j$$

I don't see how this helps though because theta is changing as is the acceleration.

 I've combined forces in the x/y direction to get:

i direction: $$\ddot{x}=\frac{-F_{Dx}}{m}$$

j direction: $$\ddot{y}=\frac{-F_{Dy}}{m}-g$$

Last edited: Oct 10, 2004
4. Oct 10, 2004

### faust9

i direction: $$\ddot{x}=\frac{-Kv^2\cos \theta}{m}$$

j direction: $$\ddot{y}=\frac{-Kv^2\sin \theta}{m}-g$$

Is this correct thus far?

5. Oct 10, 2004

### faust9

In know also that $\dot{x}(0)=100\frac{m}{s}\cos \theta,\ \dot{y}(0)=100\frac{m}{s}\sin \theta$ and $x(0)=0,\ y(0)=0$

Last edited: Oct 10, 2004
6. Oct 10, 2004

### faust9

OK, I've got this so far:

$$m\ddot x+Kv^2\cos \theta=0$$
$$m\ddot y+Kv^2\sin \theta+mg=0$$

I have to use matlab to display a graph of this using the ODE45 function. Unfortunetly, I've never used matlab before so this is quite a problem.

From whay I understand, I have to set up an m-file containing a vector where all of the ODE functions are stored.

I think I need a vector like this : v={v(1),v(2),v(3),v(4)} where:

v(1)=x
v(2)=y
v(3)=x_dot
v(4)=y_dot

I also have the following:$\dot x(1)=x(3)$, $\dot x(2)=x(4)$, $\dot x(3)=\ddot x$, and $\dot x(4)=\ddot y$.

thus:
$$m\dot x(3)+Kv^2\cos \theta=0$$
$$m\dot x(4)+Kv^2\sin \theta+mg=0$$

which I can solve for x_dot (x(3)_dot) and y_dot(x(4)_dot) but I'm not entirely sure if I need to:
$$\dot x(3)=\frac{-Kv^2\cos \theta}{m}$$
$$\dot x(4)=\frac{-Kv^2\sin \theta}{m}-g$$

I am so lost with this right now. What exactly do I do with the above to get a plot of projectile motion in terms of x and y and a plot of velocity in terms of x and y? Is there anyone out there who can help?

Last edited: Oct 10, 2004