Can you help me with a question on projectile motion?

well if it was being launched from the ground i would say its initial velocity would be 0. And its vertical velocity when it reaches its peak is also going to be 0. I dont have enough info to find its vertical acceleration and its horizontal velocity.

Deceleration is gravity 9.8m/s^2    
 U=Initial Speed
V=Final Speed
A= Acceleration
S= Displacement (Distance)

Use V^2 = U^2 + 2AS            

So  0 = U^2 - 19.6 (1210)
 
So U = 154m/s

Motion with constant acceleration is covered by two equations:
distance x = X0 + V0 * t + A * t^2 /2
velocity v = V0 + A * t
where: X0, V0 are initial distance and velocity, A is acceleration, t is time.
These two formulae are easy to remember, because velocity is a derivative of distance and acceleration is the second derivative. All other formulae are derived from these two.
You know that vertical acceleration A is -9.81m/s^2 and that at the peak height vertical velocity must be zero, so;
Vy0 * t - A * t^2 /2 = y
Vy0 - A * t = 0
Solving
t = Vy0 / A
Vy0 * Vy0 / A - A * (Vy0 / A)^2 / 2 = y
Vy0^2 / A - Vy0^2 / (A * 2) = y
Vy0^2 / (2 * A) = y
Vy0 = sq root(2 * A * y) = (2 * 9.81m/s^2 * 1210m)^0.5 = 154m/s
That's the vertical component of initial velocity. To get the initial velocity itself you need to specify how far the projectile has to land.

Questions of this type all use Newton's equations.
The position of a body is given by: D = d(0) + v(0)*t + 1/2*a*t^2
d(0) is the initial position; it is equal to 0,
we want to find v(0).
D is 1210 m.
a is the acceleration due to gravity, ~ 9.8 m/s^2.\
So Newton's equation simplifies to:
1210 = 0+ v(0)*t - 1/2*9.8*t^2.  (The minus sign in the last term is because the acceleration due to gravity is down.)

Looks like we have one equation in 2 unknowns - so what else do we know?  At the peak of it's trajectory, the projectile has a (vertical) velocity of 0.  How does that help?  Newton also derived a formula for the velocity of an object under acceleration. (It's the first derivative of the equation for position with respect to time, if you're interested in calculus.)  This equation is V = v(0) + a*t.  V, at the time we are interested in is, again, 0, so this equation becomes 0= v(0) - 9.8t.
We can solve this equation for v(0):  v(0) = 9.8t.

Substitute this into the equation for distance:
1210 = 9.8t^2 - 1/2*9.8t^2 = 4.9*t^2
Now this we can solve for t:  t = sqrt(1210/4.9)
Now we just use another "fact" we know, and use only the positive root, since we don't care what happened before we started.

And now that we know t, we can substitute into the velocity equation and get v(0) = 9.8*sqrt(1210/4.9)

And that's the answer you need.