This article was inspired by the annual Punkin Chunkin competition at Sussex County, Delaware. The aim of Punkin' Chunkin' is to throw a pumpkin as far as possible, without it being destroyed in the process. Many different methods of doing this can be used, the most powerful being air cannons. These can hurl pumpkins way over 4000 feet.

We will look at the math behind punkin chunkin. How fast does a pumpkin have to be thrown to reach 4000 feet or more? And what is the best angle of elevation?

First, let's start with a comparatively easy problem: how far could we throw a pumpkin if air resistance was not a factor? Answering this question will give us some idea as to whether air resistance is indeed important enough to need to be taken into consideration.

Consider a pumpkin-shooting cannon which can accelerate a pumpkin up to a velocity V₀ - the muzzle velocity - and suppose that the angle of the cannon's barrel to the horizontal is a.

The velocity of the pumpkin can be split into two components, its vertical velocity in an upwards direction, V_v and its horizontal velocity, V_h.

We can calculate the initial values of these two velocities using simple trigonometry: V_v = V₀.sin(a) and V_h = V₀.cos(a).

These two velocity components are independent of each other. Clearly, the horizontal component V_h is constant, as there are no forces acting on the pumpkin in the horizontal direction, in the absence of air friction.

On the other hand, in the vertical direction we have the force of gravity always pulling the pumpkin downwards to the ground. Let's call this acceleration due to gravity g. Then the vertical velocity V_v, initially upwards, decreases in proportion to the time of flight, t.

So at any particular time t, we have V_v = V₀.sin(a) - g.t.

Assuming that we are on level ground, or rather, assume that the landing site is at the same height as the cannon's muzzle, the pumpkin is at its highest point when its vertical velocity is zero, In other words at time t_m = V₀.sin(a) / g.

The path taken by the pumpkin is symmetrical about this time of maximum height. It therefore arrives at the landing site at time t =2.t_m, and the distance covered across the ground is this time multiplied by the horizontal velocity, or s = 2.V₀.sin(a) / g.V_h.

We can substitute the value of V_v into this equation, to obtain the range as s = 2.V₀.sin(a) / g.V₀.cos(a).

Using a basic trigonometric identity, sin(2.x) = 2.sin(x).cos(x), we can simplify the equation to s = V₀² / g.sin(2.a). This clearly has its maximum value when 2.a is 90 degrees, i.e. when the cannon's angle of elevation is 45 degrees.

The predicted maximum range is therefore s = V₀² / g. Taking g as 32 ft/sec², and aiming for a range of 4000 feet, we obtain V₀ = (s.g)^1/2 which evaluates to about 380 ft/sec, or about 244mph. (In metric units, take g as 9.80665m/s² and s as 1219.2m. Then V₀ is about 109.34m/s or about 394km/h.

If you've ever stuck your hand out of a car at cruising speed, you will know that we surely cannot ignore air resistance for a pumpkin, much larger than your hand, and travelling much faster through the air.