Intuition

Here's one way: Let's rotate your diagram.

Now the rocket is a cannonball!

Physics

It has a fixed acceleration "downwards" i.e. perpendicular to the vector from its firing location to its target. I drew it above as a dashed green line. Let's call that the reference horizon. (Note that this reference horizon is constant! The rocket was fired from a fixed position with a fixed position as a target.)

We know (from wikipedia) for a cannonball without air resistance, that d = v^2 * sin(2 * theta) / g, where

• d is the horizontal distance travelled (distance between firing location and target)
• v is the speed the projectile was fired at
• theta is the angle as to the horizon the projectile was fired at (angle of fire direction vector from the reference horizon)

Rearranging the equation for g gives g = v^2 * sin(2 * theta) / d.

The constant in the cannonball equation, g, is acceleration due to gravity. We can take it to mean acceleration due to rocket propulsion. It doesn't matter, sinceThat's fine too — it's still a constant acceleration in a constant direction.

Now what?

Run that equation for g when you fire the rocket. It will tell you how much to accelerate the rocket perpendicularly toward the reference horizon, in order to hit the target. Since the direction of that acceleration is constant, an orbit won't form.

Boom.