By Laura Aseltine
In a desperate attempt to find something to do for my physics news I googled “Weird News” hoping to find something I could apply my limited physics knowledge to and came across the annual event that is a physics lover’s dream known as “Punkin Chunkin.” The competition occurred on November 1-3 and a television special covering this year’s event aired on Thanksgiving day (unfortunately I was too busy eating to remember to watch). The ultimate goal is simple: how far can you chuck a pumpkin, but there are many different categories and specific rules for how you can chuck this pumpkin. Some of the categories include air, centrifugal, catapult, and trebuchet.
For my physics news I decided to look at the air cannon, which uses compressed air to propel the pumpkin (pretty self explanatory). The official Punkin Chunkin website lists the rules as follows:
1. Pumpkins must weigh between 8 & 10 pounds.
2. “Compressed air only"
3. Pumpkin must be loaded before Pressurizing vessels, and an Official must see you load it.
4. All Air inlets on vessel must have a Check Valve Installed.
5. Horn or sound device must sound when firing down range for safety of spotter on the field.
This years winner of the air cannon category was a team by the name of American Chunker Inc who shot their pumpkin a world record distance of 4,694.68 ft. I decided to calculate what the initial velocity of that pumpkin must have been when it left the cannon.
What I know/Assumptions
· I found online that the cannon is 120 ft long
o 120 ft x 0.3048 m/1 ft = 36.576 m
· I assumed the cannon is at a 70° angle
· Calculated the initial height of the pumpkin
o Sin70 x 36.576 m = 28.30 m
· Distance traveled = 4694.68 ft
o 4694.68 ft x 0.3048 m/1 ft = 1430.94 m
- To find the initial velocity I need to break it up in into x and y components
o V0x= V0 cos70
o Voy= V0 sin70
- First I calculated the time it took to reach the maximum height
o Vf= V0 + at
o 0 = V0 sin 70 – 9.8t
o t = 0.0793 V0
- Then I calculated the maximum height it reaches
o Vf2 = V02 + 2a(Δy)
o 02 = (V0 sin70)2 + 2 (-9.8) Δy
o -(V02 0.599) = -19.6 Δy
o 0.031 V02 = Δy
o max height = (V02 0.031) + 28.3
- Then I calculated the time it would take for the pumpkin to fall from the maximum height to the ground
o Δy = V0t + ½ at2
o – ((V02 0.031) + 28.3) = 0 + ½ (-9.8)t2
o √.0063V02 + √5.776 = t
o t = .0793V0 + 2.403
- So the total time the pumpkin was traveling is…
o t = 0.0793V0 + .0793V0 + 2.403
o t = 0.1586V0 + 2.403
- I then used this time to calculate the initial velocity in the x direction
o Δx = V0xt + ½ at2
o 1430.94 = V0 cos70 (0.1586V0 + 2.403) + 0
o 1430.94 = 0.100V02 + 1.522V0
o V0 = 119.38 m/s
- I then converted this velocity to miles per hour
o 119.38 m/s x 1 mi/1609.34 m x 3600s/hr = 267.05 mph