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 !