Thanks heaps, man. I just did the question myself, though, while I was waiting for a reply! lol
WHOOOO! So stoked about it, too. I'm not sure how everyone else went with it, but I'm posting my solution up anyway!!! Maybe that'll bring me down from my high!
The Question is 7(b)iii. from the 2003 paper; check it out before reading my sol'n.
dmax occurs when α = π/4
This holds for all h ≤ H - S (and ∴ all d ≤ the value of d when α = π/4, h = H - S)
When α = π/4, h = H - S, and ∴
( v2sin2α )/2g = H - S
v2(1/√2)2 = 2g(H - S)
v2 = 4g(H - S)
For values of d ≤ the value of d when α = π/4, h = H - S,
v2 ≤ 4g(H - S)
(Since both h and d are decreasing, and α remains the same, v, and ∴ v2 must decrease)
And d = ( v2sin2α )/g = [v2sin(2.π/4)]/g = v2(1)/g = v2/g
But, for values of d ≥ the value of d when α = π/4, h = H - S,
α must decrease and v must increase to sustain both a constant h and an increase in d
(The fixing of h to H - S is based upon the fact that, ceteris paribus, the closer α is to π/4, the greater will be the range of the projectile - I didn't include this in my proof lol)
i.e. v2 ≥ 4g(H - S)
Bring the constant h into the equation for d (where v and α now vary):
d = ( v2sin2α )/g =
[2( v2sinαcosα )]/g =
( v2sin2α )/2g x 4cosα/sinα =
4(H - S)cotα =
4√[(H - S)2cot2α] =
4√[(H - S)2(cosec2α - 1)] (From 1 + cot2x = cosec2x)
But ( v2sin2α )/2g = H - S
i.e. v2/2g = (H - S)/sin2α = (H - S)cosec2α
So, d becomes:
d = 4√[(H - S)(H - S)(cosec2α - 1) - (H - S)2] =
4√[(H - S)(v2/2g) - (H - S)2]
Q. E. D.
Now, that is one f**king nasty 3 Unit Question! Enjoy!
