If there isn't any air resistance, then the maximum height will be reached at the halfway point between O and Q. (this isn't too hard to prove, and it should also make sense) This means that you can find the height from a right angle triangle. (which is half of the equilateral triangle OPQ Leodip mentioned) All you need for that is Pythagoras' Theorem, which is D^2 + E^2 = F^2 (using D, E, F so as not to confuse with other labels) F is the length of the line OP (or OQ) and I call that X. One of D and E is the unknown height P and the other is half of X. So you have

P^2 + (0.5X^2) = X^2

and you end up with

height = P = X sqrt(3) / 2

Using the fact that when the projectile is at point P, it has zero vertical velocity, you can take the basic equation

distance = initial velocity * time + 0.5 * acceleration * time^2

The initial velocity, as I just said, is zero, so you can ignore that term, and solving for time, you will get that

time = sqrt (2 * distance / acceleration) (note that this is the time to fall from P to Q)

The distance is the height I just figured out, sqrt(3) / 2, and the acceleration is the acceleration of gravity, since this is only considering the vertical height, velocity,

and acceleration. So if you substitute in the height from above, the time that the projectile takes to fall from P to Q is:

t = sqrt ( sqrt(3) X/g )

This is only half of the time that the projectile is in the air, though, because it spent an equal amount of time going from O to P. So the entire journey's duration, T, is double this time, which is

T = 2t = 2 sqrt (sqrt(3) X/g)

The horizontal velocity Vh can be calculated from distance = velocity * time, or velocity = distance / time

Vh = distance / time

The distance is defined by the problem as X. The time is T.

Vh = X / (2 sqrt (sqrt(3) X/g) )

The vertical distance Vv can be calculated from velocity = distance / time as well.

This time, the distance is the height P calculated earlier.

Vv = distance / time = ( X sqrt(3) / 2 ) / ( 2 sqrt (sqrt(3) X/g) )

The velocity, v, that Leodip wants to solve for is found by taking both components into consideration. The equation is:

v^2 = (Vh)^2 + (Vv)^2

v^2 = ( X / (2 sqrt (sqrt(3) X/g) ) )^2 + (( X sqrt(3) / 2 ) / ( 2 sqrt (sqrt(3) X/g) ))^2

Because this is butt ugly, doing some algebraic simplification gives

v^2 = 7xg / (16 sqrt(3) )

v = sqrt ( 7xg / (16 sqrt(3) ) )

That's one of the things Leodip wanted to find.

I'll have to think on the angle later; I don't have time to finish it right now. Feel free to ask questions or point out mistakes, but it'll be a few hours at least before I check in again.