Path of a projectile from a cannon

[kLPantera]

Homework Statement

Part 1

We fired a cannon 0 degrees to the horizontal line and had to calculate V_o. We then measured Delta x and Delta Y.

Part 2

We were then given a degree in which to fire the cannon at, which was 12 degrees. We then had to calculate the projectile's path through 3 rings and then hit a target.

We we used were what we got from part 2 (V_o, Delta x, and Delta y).

Data:
Delta y = -0.87m (from a counter top to the ground)

Calculated Data:

Vi_x = 4.02 m/s
Vi_y = 0.85 m/s
V_o = 4.11 m/s

Known Data:

a_x = 0 (we are operating under the scenario that there is no air resistance)
a_y = -9.8 m/s²

Homework Equations

Delta y = Vi_yt + (1/2)gt²
Delta x = Vi_xt + (1/2)at²
V_fy = Vi_y + gt
Theta = tan^-1 = (Vi_y/Vi_x)

The Attempt at a Solution

First I calculated time:

Delta y = Vi_yt + (1/2)gt²
-0.87 = (0.85)t + (1/2)(-9.8)t²
t = 0.5163 seconds ( I used quadratic formula)

Second I calculated Delta x:

Delta x = Vi_xt + (1/2)at²
Delta x = (4.02)(0.5153)
Delta x = 2.07

There are three rings placed at equal intervals throughout Delta x so:

x₁ = 0.5175(ring 1)
x₂ = 1.035 (ring 2)
x₃ = 1.5525 (ring 3)
x₄ = 2.07 (target)

Time to reach each target:

Ring 1: 0.129075 seconds
Ring 2: 0.25815 seconds
Ring 3: 0.387225 seconds
Target: 0.5163 seconds

Height of each ring:

Delta Y₁ = (0.86)(0.129075) + (1/2)(-9.8)(0.129075)² + 0.87 = 0.898
Delta Y₂ = (0.86)(0.25815) + (1/2)(-9.8)(0.25815)² + 0.87 = 0.76
Delta Y₃ = (0.86)(0.387225) + (1/2)(-9.8)(0.387225)² + 0.87 = 0.464
Delta Y₄ = (0.86)(0.5163) + (1/2)(-9.8)(0.5163)² + 0.87 = 0.008

Angle For Each Ring:

Ring 1: V_y = (-9.8)(0.129075) + 0.85 = -0.41493
theta = tan^-1 = (0.41493/4.02) = 5.89 degrees

Ring 2: V_y = (-9.8)(0.25185) + 0.85 = -1.67987
theta = tan^-1(1.67987/4.02) = 22.67 degrees

Ring 3: V_y = (-9.8)(0.387225) + 0.85 = -2.944805
theta = tan^-1(2.944/4.02) = 36.21 degrees

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With these calculations my lab group and I attempted to shoot through Ring 2 and hit the target. Our ring 2 fell short of the projectile, however the projectile still hti the target close to the bullseye. We also have to shoot through ring 3.

Could someone help?

Much Appreciated.

 

[zgozvrm]

Is there a specified distance from the cannon for each of the 3 rings?

 

[zgozvrm]

Also, what were [itex]\Delta[/tex]X and [itex]\Delta[/tex]Y for the projectile when fired at 0 degrees?

 

[kLPantera]

Fired at 0 degrees:

Delta x:

For 1 click: 1.28 meters
For 2 clicks: 1.68 meters
For 3 clicks: 2.28 meters

Note* clicks means how much we compressed the spring within the cannon

Delta y = 0.87 meters

Each of the three rings are set at equal intervals of 1/4 of delta x.

 

[zgozvrm]

Calculated Data:

Vi_x = 4.02 m/s
Vi_y = 0.85 m/s
V_o = 4.11 m/s

How did you come up with these values?
(This appears to be the beginning of your problems).

 

[zgozvrm]

Fired at 0 degrees:

Delta x:

For 1 click: 1.28 meters
For 2 clicks: 1.68 meters
For 3 clicks: 2.28 meters

Note* clicks means how much we compressed the spring within the cannon

Delta y = 0.87 meters

Each of the three rings are set at equal intervals of 1/4 of delta x.

How many clicks did you use when you fired the cannon at 12[itex]^\circ[/tex]?

 

[zgozvrm]

Second I calculated Delta x:

Delta x = Vi_xt + (1/2)at²
Delta x = (4.02)(0.5153)
Delta x = 2.07

...but you already calculated [itex]\Delta[/tex]x:

Fired at 0 degrees:

Delta x:

For 1 click: 1.28 meters
For 2 clicks: 1.68 meters
For 3 clicks: 2.28 meters

 

[kLPantera]

First we fired at 0 degrees, which we then used to find V_o

Vi_x and Vi_y were found after he gave us the angle of degree to fire it at.

For Delta x we had a Delta x for when it was 0 degrees. We then were given an angle to fire at. We have to find the Delta x for when it will be fired at 12 degrees.

We used 2 clicks when firing at 12 degrees.

 

[zgozvrm]

So, for 2 clicks at 0[itex]^\circ[/tex], you had a [itex]\Delta[/tex]x of 1.68 meters.
Also, at 0 degrees, Vi_y is 0 m/s
Therefore, using [itex]\Delta y = V_{iy} t + \frac{1}{2}a_y t^2[/tex], we get
[itex]-0.87 = \frac{1}{2}(-9.8)t^2[/tex]
Solving for t, we get 0.421 sec.

Now, using [itex]\Delta x = V_{ix} t[/tex], we get (for 2 clicks):
[itex]1.68 = V_{ix} (0.421)[/tex]
Solving for [itex]V_{ix}[/tex], we get 3.99 m/s

Therefore [itex]{V_o}^2 = {V_{iy}}^2 + {V_{ix}}^2[/tex] and, thus [itex]V_o[/tex] = 3.99 m/s


Use this value of [itex]V_o[/tex] for any angle of projection using 2 clicks.

 

[kLPantera]

Thanks again!