# Vector forces on projectile motion

1. Sep 20, 2008

### GTOzoom

I am doing a fairly large projectile motion problem. I am trying to describe the act of throwing a ball from an elevated point at an upward angle into a lower point, while taking into account the affects of ball spin, air resistance, etc.

First, I have a question about the act of throwing a ball. When you throw a ball, you form a lever with you elbow, so the motion of throwing the ball makes an arc. When you release the ball does the ball continue tha arc, or does it start moving with a vector magnitude when you release it?

Next I have a question about my equations. The y-component of my vector forces at this point is: sin(a)Vn(m)-mg where a is the angular component of the vector, V is the magnitude of the vector, m is the mass of the object and g is gravity. The x-component is: cos(a)Vn(m). I don't know if my equation is right however, being that the result is going to be an arc, doesnt that mean that I would need to find some way to calculate the tangent line of the previous point in the arc to clculate the next point? I guess what I am asking is how to I take my x and y vector components, apply gravity to them and turn them into a singular equation for my arc?

Also, I need to know how to do this because later on I need to apply air resistance, and the spin of the ball to the equations

oh, also, because it forms an arc, does the ball have angular momentum? I don't think it does because it is just vector forces acting on the ball at this point, I understand that once I add spin it will, but the arc of a non-spinning ball only has linear momentum right?

2. Sep 20, 2008

### Perillux

The ball always wants to travel in a straight line, the only reason it travels in an arc is because there is a constant acceleration pulling it toward the center of the arc.
When you release the ball it will travel along a straight path tangent to the curved path it was on originally. Of course then gravity will cause it to follow another curved path.

For your second question I think your equations are wrong. Projectile motions are independent of mass. If you need help doing the calculations then I might be able to help.
It sounds a little like you are trying to make a program to do this and I recently wrote some projectile motion code. However, it does not include air resistance or spin.

and yes it's all linear.

3. Sep 20, 2008

### atyy

Every linear motion problem is also an angular momentum problem, because

(angular momentum) = (linear momentum) X (perpendicular distance about a point)

where the point can be any point. For any problem which can also treated as purely linear, this is just a correct but stupid way of looking at it. But when you really need to include angular momentum, it's sometimes helpful to know about this formality.

4. Sep 21, 2008

### GTOzoom

Thanks for the reply's...what i'm actually trying to do is create a set of equations to solve for the optimal beruit shot. In the end what I want to do is plug in 2 shots(minimum distance to cup1, maximum distance to cup1) into the formula's and solve for v, then take V2-V1 and solve for the angle of the initial vector and use a computer to maximize that function, thus giving the best angle to maximize the difference between the velocities of the 2 extreme successful shots.

This also means that when my equations are done, I will take the final angle of descent an d make sure the ball clears the front of the cup, and stays within the balls cg down from the back of the first cup.

So about the equations. I realize that projectile motion doesn't involve mass, but air resistance is a huge factor due to the light weight of a ping pong ball, I just whipped together those momentum formula's figuring that I would need them down the line. I researched more into the topic and I found these sets of equations: http://www.grc.nasa.gov/WWW/K-12/airplane/flteqs.html

Now according to this website, you need to apply differential calculas to those sets of equations. But I think that there is an easier way to apply air resistance,which might not be quite as accurate, but is accurate enough. I was thinking that you could find the terminal velocity of the ping pong ball, making mg=Cd, and applying that drag coefficient opposite the tangent line of the arc. It makes sense, but i'm having trouble actually applying that reasoning mathematically.

Next is spin, I have been reading up on it and it sounds like the arc of the shot changes with spin due to lift and drag. I have the formula for the ideal lift of a smooth sphere, which i got again from nasa's site. Lift=4/3(4pi2b3spV)
s=rotation
p=density
V=velocity
the url is: http://www.grc.nasa.gov/WWW/K-12/airplane/beach.html

this looks very similar as well: http://www.grc.nasa.gov/WWW/K-12/airplane/foil2b.html

Its the equations that I'm having trouble with, its been about 3 years since I took physics and I recently decided to go to college to major in physics, so I'm trying to hop back on board by giving myself an interesting problem to solve. Thanks for all the help guys!

5. Sep 22, 2008

### GTOzoom

I was trying to come up with a simple equation that doesnt take into account air resistance, spin, or anything and I ran into some problems.

my equation for the x-axis(horizontal distance) is :x=cos(a)vt
my equation for the y-axis(vertical distance) is :y=sin(a)vt+(0.5*(-9.8))t^2

I then solved for t in the first equation and plugged it into the second equation, giving me:

y=(sin(a)v)(x/(cos(a)v))+((1/2)(-9.8))(x/(cos(a)v))^2

The arc looks about right, but when I calculate the tangent line of point (o,o) (angle=60, velocity = 10) I get a line with an angle of about 17 degrees. Why isn't this formula working?

6. Sep 28, 2008

### GTOzoom

I'm almost done with my flight equations(neglecting spin) this is what i have:
Vt=terminal velocity
Vo=initial vert. velocity
V= local vert. velocity
g=gravity
t=time
Uo= inital hor. velocity

sum of Y forces = ((Vt^2)/(2g))*ln((Vo^2+Vt^2)/(V^2+Vt^2))
sum of X forces = ((Vt^2)/g)*ln((Vt^2+gUot)/(Vt^2))

solving the second equation for (Vt^2)/(2g) we get : x/(2ln((Vt^2+gUot)/(Vt^2)))
Plugging that into the first equation leaves us with:

y = x/2ln((Vt^2+gUot)/(Vt^2)) * ln((Vo^2+Vt^2)/(V^2+Vt^2))

If you notice there is still V(local velocity) in the equations, I need to get that OUT. So...how to I calculate the local velocity and then plug it into that equation? any help would be very much appreciated.

oh btw, the other equation I posted is correct, I was in radians.