Punkin Chunkin

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   V_0x= V₀ cos70

o   V_oy= V₀ sin70

-       First I calculated the time it took to reach the maximum height

o   V_f= V₀ + at

o   0 = V₀ sin 70 – 9.8t

o   t = 0.0793 V₀

-       Then I calculated the maximum height it reaches

o   V_f² = 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 = V_0xt + 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