A bow stores elastic potential energy in the flexible ash of the wood of the bow and the stretched string.

The English archers won the Battle of Agincourt in 1415 because the range of the better-made and longer English bows was greater than the bows used by their French enemies. They stored and released energy more efficiently.

They could probably count on a release speed of 100 feet per second = 3000 cm/s = 30m/s and they knew that a 45 degree angle gave maximum range.

So 30 cos 45^{0} = horizontal speed = 21.2m/s. This stays the same because a=0 in the horizontal plane.

In the vertical plane, we have to use the equations of motion (constant a). With 30sin45^{0} as u, and acceleration = g (9.81ms^{-2}) this yields t = 2.16s going up, PLUS the same coming down so the arrow’s total time in the air = 4.32s.

So, range =21.2×4.32 = 91.6m…and vertical height reached is 33.2m as long as air resistance is neglected.

Now, let’s take this apart…

At the top of its flight, what kind of energy does the arrow have? Estimate arrow mass at 0.1kg (is this reasonable?)

A: Using kinetic energy in the horizontal direction (22.47J) plus GPE at maximum height (32.57)

**Estimated energy 55J**

Where does this energy come from? Think about the area under a F/x graph for a spring or similar where Hooke’s Law is obeyed…

A: the energy stored in the stretched bowstring. Let’s assume it obeys Hooke’s Law.

Using our calculations so far, we can now estimate the maximum pulling force the archer would have to use. Suppose the bow is drawn a distance of 0.5m

With quite a lot of ifs and buts – you can see where the approximations are – the archer would have to pull with a maximum force of 55/0.5N or 110N. This would be like holding a mass of 11kg just off the ground. With one hand…

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