The Hardest Problem

The following Question, 16(b), is deemed to have been the hardest problem on the Maths Extension 2 paper of this year’s HSC (Higher School Certificate), which I think is the Australian Equivalent of the Leaving Certificate. You might find a question like this in the Applied Mathematics paper in the Leaving Certificate actually. Since it covers topics I’ve been teaching here in Maynooth for first-year students I thought I’d share it here.

I don’t think it’s all that hard really, probably because it’s really a physics problem (which I am supposed to know how to solve), but it does cover topics that tend to be treated separately in school maths (vectors and mechanics) which may be the reason it was found to be difficult.

Anyway, answers through the comments box please. Your time starts now.


6 Responses to “The Hardest Problem”

  1. John Peacock Says:

    It’s a nice problem, more like the sort of thing I remember getting in advanced maths A-level in the 1970s, and probably harder (or at least terser) than what kids today are typically facing. I have to say, I still find it an enormous amount of work to generate a new cute problem of this sort. Usually if I think of an interesting question the solution turns out to be ugly.

  2. Going to need a hint here. Tried scalar product set to zero but its not helping me find the angle of projection.

  3. Ok, I found the scalar product of r(t) and v(t) and equated this to zero. This gave me t = 3usin(theta)/2g +/- u/2g rt(9sin^2(theta)-8). As we need two solutions for t, the root must be real meaning theta > 70.5 degrees. I am guessing the condition is 70.5 < theta < 90. However, there appears to be two solutions for theta =90. I cannot visualise how the velocity and displacement vector could be perpendicular when it is projected vertically up.

    Anyway, there are two distinct and real solutions for each angle because rt(9sin^2(theta)-8) is always < 3sin(theta) giving all positive solutions for t.

    So where did I go wrong?

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