Previous answer is half right. That will give you the vertical component v_y of velocity. Now you need a way to get the horizontal component.
Figure out how much time the rocket is in the air, given that vertical velocity. Then figure out what the horizontal velocity is, given that it travels 300 km in that time.
The time from initial launch to peak is the time to reach velocity 0 starting at v_y: t = v_y/a. The total time in the air is twice this, 2*v_y/a. So the horizontal velocity is 300*10^3 m / [2*v_y/a].
Once you have horizontal and vertical velocities, you can find the magnitude and direction of the resulting vector.
