how to do part iii)
This is like the poster child of disgusting questions for Extension 1. It is the 2003 HSC Projectile Question; and probably the most brutal projectile question I know of
I haven't done this one in ages, so this will be interesting.
We need to consider
two cases here. When \(\alpha=\frac{\pi}{4}\), \(d\) takes its maximum value (this isn't something you need to prove). So, let's roll with that in Part (ii):
Now this is valid so long as the height of the projectile doesn't cause it to hit the roo, ie, \(\text{Height} < H-S\). Let's go to Part (i):
)
So that's the first half of the answer done!
Now we consider other values of \(\alpha\), those which will yield a smaller value for distance. These correspond to v^2\lge4g(H-S). The derivation for this is similar, but use some common sense: What happens to the hright of a projectile as you increase its launch angle?
Now let's tie in to Part (ii):
But from Part (i), that first bit is (H-S):
\cot{\alpha}=4\sqrt{(H-S)^2\cot^2{\alpha}}=4\sqrt{(H-S)^2(\csc^2{\alpha}-1)} \quad(A))
Applying the square root and then fixing what's inside to match is just to get closer to the form of the answer, and also to apply that pythagorean transformation. Continuing:
\implies\csc^2{\alpha}=\frac{v^2}{g(H-S)})
Substitute that into the line A, and we have our answer. This is a BRUTAL question, be sure to let me know if this makes sense, because I was very quick with algebra! 
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