Launching a Small Ball: Targeting Rings & Floor

[username111]

Homework Statement

i'm launching a small ball from a launcher (on a table) at 55 degrees into rings (at .25, .50, .75, 1) and to a target on the floor.

height from the floor to the launcher: 1.015m
angle: 55
velocity: 6.62 m.s

i have to find how far each ring will have to be from the launcher, how high the ring stands need to be, and how far the target on the ground has to be (to go from the launcher through each ring, and then it the target.)

http://img168.imageshack.us/img168/1156/img005na3.jpg <-- it was a bit big to put in as an image

Homework Equations

y=(Vy)(t)+1/2gt^2
x=(Vx)(t)

The Attempt at a Solution

http://img135.imageshack.us/img135/544/img006fp1.jpg <-- too big to put in as an image

for the distance the rings have to be i used x=(Vx)(t).
i don't think i set this up right though.

how i did it:
first ring
x=(Vx)(t)
x=(.25)t
x=(.25)(1.107s)
x=0.277m

second
x=(.50)(1.107s)
x=.5535

the rest are in the picture.

and to find the height of the ring stands i used max height.

center ring (second one)

t=1.107s/2
t=.5525s

y=(Vy)(t)+1/2gt^(2)
y=(5.423 m/s)(.5535s) + 1/2(-9.8 m/s)(.5535s)^2

first ring height
t=.5535s/2
t=.27675

then i put it in the equation and got 1.13m.

would this be the height from the table or the ground? I'm pretty lost and this is a major grade so any help would be appreciated. i need some guidance in the right direction.

 

[chrisk]

It appears you are inserting the relative postition of the ring into Vx, the velocity of the ball in the horizontal directions. Clarify what the .25, .5, .75 and 1 values represent.

 

[LowlyPion]

What exactly are you trying to do?

Place the rings at 25%, 50%, 75% and 100% of flight time?

Or .25m, .5m, .75m, 1m ?

I don't understand your answer in times at all.

Neither do I see you using the horizontal velocity component 3.79 m/s to determine the horizontal spacing.