How do I calculate Initial, horizontal, and vertical velocity with only angle, time, and max height?

If you have teh time and the max height (I assume time is the total time of flight), then you can estimate the inital speedin y by "working backwards".  Max height occurs 1/2 way thorugh the flight and it takes the object just as long to fall from max height as to rise to max height.  So treat this as a falling objecty and find the initial y speed from:

  vymax = vy0 -g(t/2)  where t/2 = 1/2 teh total trip time and vymax = 0 since the object stops moving in y at max height

  0 = vy0 - 1/2 gt --->  vy0 = 1/2 gt

Now you know that vy0 = v0 sin(q)  where v0 = initial velocity and q = launch angle  so

  v0 = vy0/sin(q)

Finally vx0 = v0*cos(q) = vy0*cos(q)/sin(q) = vy0*cot(q)

If the launch velocity is V at an angle w the
vertical component = Vsin(w)
horizontal component = Vcos(w)
When it reaches max height speed = 0 instantaneously.
H = Vsin(w)T - (1/2)gT^2  .............. [From s = ut + (1/2)at^2]
V = {H + (1/2)gT^2}/(Tsin(w))
So you now also have Vsin(w) and Vcos(w)

Let's work out an example.

Your projectile attains a maximum height of 1,100 meters, and is launched at an angle of 34 degrees above the horizontal.

The total flight time is 29.966 seconds.
Sqrt(1100 / 9.8 * 2) * 2 = 29.966

The initial vertical velocity is 146.833 m/s.
Sqrt(1100 / 9.8 * 2) * 9.8 = 146.833

The initial horizontal velocity is 217.690 m/s.
146.833 / tan(34) = 217.690

The initial velocity is 262.581 m/s.
Sqrt(146.833^2 + 217.690^2) = 262.581