Projectile Motion Questions Please !!

Number 1)
A particle is released at Vms^-1 and at A° from a point 15 metres above ground level. It just clears a fence 30 metres away and 26.25 metres high. What is V and what is the angle?
Working out:
So far, ive managed to get a cartesian equation like this:
y = (-5x^2/V^2)(1+tan^2A) + xtanA + 15
Now even if i DO substitute the two co ordinates for the fence (30,26.25) --> 26.25 = (-4500/V^2)(1+tan^2A) + 30tanA + 15 , i am still left with TWO variables, the velocity and the angle !!! So what do i do ???

Number 2)
There are two walls with same height 7 metres. One is 7 metres away and one is 14 metres away from the projectile launch (so 7,7 and 14,7).
a) Prove if A is the angle, then tanA = 3/2
b) If the walls are H metres high and distance B metres and C metres from the point of projection, prove that tanA = [H(B+C)]/BC
Workint out:
No idea! Please help me !

As far as I can see there has to be two variables in your first question. If you shoot really really hard and with a low angle it will hit that point of yours, but you can also hit it if you shoot really really high, and it will hit it on it's way down again.

But for one angle, there is only one speed that goes with that angle.

ANSWER: The ANSWER is 25ms^-1 and 37° . But HOW do i get it?