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Going to reiterate this, I still believe it to be beneficial that it is stickied

Re: HSC 2012 Marathon :) Very nice way of looking at it, I did it by treating (t+T) as a single parameter and found equation of projectile B then solved from there. And I too must get back to the hexagon, post a solution if you find it out :)

Re: HSC 2012 Marathon :) Oh of course, sorry for that, yes the particles collide at maximum height. Sorry for the inconvinience

Re: HSC 2012 Marathon :) $A projectile is fired from the origin with velocity V at an angle $ \theta $ to the horizontal, the projectile is fired such that the maximum height is h. T seconds later a projectile is dropped h metres above the point where the first projectile is to reach its...

Re: HSC 2012 Marathon :) Very nice, new question will come up soon.

Re: HSC 2012 Marathon :) Nope, your method is correct, I did it and it yielded my results (45, 135) Did you end up with: \frac{8\sin^2 \theta-4}{\sin^2 \theta \cos^2 \theta}=0 \Rightarrow \sin^2 \theta=\frac{1}{2} \\ \\ sin \theta=\frac{1}{\sqrt{2}}

Re: HSC 2012 Marathon :) Another question I made up: $A projectile is fired from the origin with velocity V, under the conditions of gravity g and no air resistance. The projectile is fired such that the maximum height is h, and the horizontal distance from the origin to the maximum height is...

Re: HSC 2012 Marathon :) To do math notation you type normal latex code within the tags [.tex] and [./tex] (without the dots) Also for the hexagon question I havent quite got the answer yet but Im quite close, I feel like I need to somehow prove that BP=BC since everything falls nicely from...

SOLUTION Posting solution since no one would learn if they didnt know :)

Just know basic algebra and the definition of first principles and you should be good to go, on a 2U level they probably wont ask stuff involving substitutions (unless they give it to you), but who knows maybe at the start of Q16 a decent 3 marks to a hard first principles question.

Re: HSC 2012 Marathon :) Ah no one answered v which made the thread die, asianese answered the other ones but you can do them, its a good excercise I reckon :)

Re: HSC 2012 Marathon :) Here is the solution for just v, or at least how I did it: http://i981.photobucket.com/albums/ae294/sy08071996/Image10.jpg I dont want to leave this question unanswered lol. Will make another question if the demand is high enough (Keep dyeing marathon alive!)

+1

I forgot to write my reasoning for this, but if the situation is like this, are we allowed to assume that the stationary point is a maximum since there is no minimum area (A=0 min) Am I allowed to say this and not differentiate? Such a pain to do it again

You are correct, and my method is correct I made a wrong calculation in the calculator, recalculated and got what you got :)

Oh thats true oops. Then yeah obviously yours is correct then (dem silly mistakes)

Did you add up the sector and the two segments? If so then I may have made a silly mistake

a) Call the centre O (for part iii) Dotted lines are sight lines. b) 2 lookout towers needed for whole pitch c) Refer to diagrams S refers to sector Seg refers to segment \triangle ABC $ is equilateral$ \\ \therefore \angle AOB=\frac{2\pi}{3} \\ \therefore $(By cosine rule)$ \ \...

I changed it lol, you should probably delete your post too so it doesnt give it away (I gave it because it would be more 2U friendly)