You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x).
To solve this problem it is necessary to apply the kinematic equations of motion.
By definition we know that the position of a body is given by
Where
Initial position
Initial velocity
a = Acceleration
t= time
And the velocity can be expressed as,
Where,
For our case we have that there is neither initial position nor initial velocity, then
With our values we have 
, rearranging to find a,
Therefore the final velocity would be
Therefore the final velocity is 81.14m/s
I don’t think I’m right but I want to say 500 m/s
