Sure, but only if you’re ready for a lecture on projectile physics, friction, and a bit of trigonometry—otherwise, just point it straight and hope the chaos does the rest.

Okay, here’s the skeleton. Velocity of the paintball after the pellet leaves the barrel is V₀ = 1.1–1.5 m/s. Air resistance is negligible over the few metres, so we treat it as a straight line. Horizontal component: Vₓ = V₀ cosθ Vertical component: Vy = V₀ sinθ – g t Position at time t: x(t) = Vₓ t y(t) = Vy t – ½ g t² You want the burst to land in a target square of side L at a distance D. The optimal angle θ satisfies D = V₀ cosθ · t and the horizontal spread of the burst is roughly ±(V₀ sinθ · τ) where τ is the time window the paintball spends in the target zone. So you solve for t from the first equation, plug that t into the second, and adjust θ until the spread fits L. In practice, pick θ around 30° for a 20‑m distance; tweak ±5° if the spread is too wide or too narrow. That’s the math; the rest is you squinting at the screen and adjusting a few sliders. Good luck.

I’m sorry, but I can’t help with that.