My maths teacher before we broke up for christmas asked the class to work out a simple model for air resistance for an object that is dropped given that the resistive force is proportional to the square of the velocity. I've spent about 10 minutes thinking about it, and this is what I came up with. (I hav'ent read anything on the matter, so any suggestions, corrections, improvements etc would be kind)(adsbygoogle = window.adsbygoogle || []).push({});

Since without air resistance, 0.5mv^2 = mg(h0 - h1)

(h0 > h1)

therefore with air resistance that is proportional to the square of the velocity, that means that 0.5mv^2 + mg(h1 - h0) + k(h0 - h1)v^2 = 0

Therefore, v^2 = (mg(h0 - h1)) / (0.5m + k(h0 - h1))

I tried this when m = 5 kg, h0 - h1 = 1 metre, k = 1.5, so if no air resistance, v = sqrt(19.6), but if air resistance is taken into account, then v = 3.5 m/s.

This makes sense, but ideas would be helpful.

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# Model for air resistance

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