Projectile angle of approach and height of target?

[kLPantera]

Homework Statement

For my lab, I need to calculate the height of the ring stands so the projectile can pass through them; as well as the angle the projectile will come at them.

I have calculated/measured the following things:

Muzzle Velocity: 4.11m/s
Vi_x= 4.02m/s
Vi_y= 0.85m/s
Delta Y: 0.87 (since the cannon we are using is being shot off the counter top, it will be -0.87 when calculating)
Delta X: 2.07 meters
t = 0.5153 seconds
a_x= 0
a_y= -9.8m/s

What is given:

Angle the cannon is shooting at: 12 degrees
I also calculated the time and distance it will take for the projectile to reach each of the ring stands. There are 3 ring stands and they are placed at the same intervals

Homework Equations

I don't know how to find the height I need for the ring stands (3 of them) or the angle the projectile will come at them. At least I am unsure which formula to use.

 

[Runaway]

To determine the height, use the equation Y=1/2 * ay * t^2 + Viy * t + Delta Y, using the time you calculated for the projectile to pass through each point. (Note, this will give you the point the projectile will pass through, so set the ring stands so that the center of the opening is at this height by adding the radius of

To find the angle, first you need to find the X and Y components of the ball's velocity, its x component should be Vix, and you can calculate its Y component with Vy= ay * t + Viy. Then all you need to do is use trig. to find the angle of the resultant vector: theta=Tan^-1(Vy/Vx)

 

[kLPantera]

radius of...? what?

When I use the equation Y = (1/2)ayt²+Vi_yt + Delta Y.

I calculated that the first ring stand is 0.5075 meters away from the counter we launch the cannon off of, and that it takes0.128825 seconds for the projectile to reach it.

So what I did was:

y = (1/2)ayt²+Vi^yt + Delta Y
y = (0.85)(0.128825) + (0.87)
y = 0.97

Is it possible for a cannon with a muzzle velocity of 4.11 m/s and a Vi_y of 0.5 and a Vi_x of 4.02 m/s to have the first ring stand that high?

 

[Runaway]

I'm saying that you have to raise the ring stand above the height you calculated so that the projectile will go through the ring rather than hit the top of it, and to give it the best chance of getting through the ring, you want the projectile to go through the center. So you need to raise it by 1/2 its diameter.(see below)

.5 * a_y * t² + vi_y * t + Delta Y = Y_t
.5 * -9.8m/s² * (0.128825s)² + .85m/s * .128825s + .87m = 0.8982 m
It looks like your math is a little bit off, but yes, it is possible.

 

[kLPantera]

Oh I understand now. However there is a slight problem, we are not told the radius or diameter of the ring on the ring stand. The only measurements we have is the Delta Y, and degree of cannon fire.