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I know $a=F/M$ and that if you launch straight up you have to subtract the rocket's weight from the thrust, but what about at an angle? At any angle there's a rearward component that is sine theta of the weight acting against the thrust, but I'm unsure of how that affects the rocket's ascent.

A rocket with mass of $5000~\rm{kg}$ and a thrust of $100~\rm{kN}$ would have a vacuum acceleration of $20~\rm{m/s}$ and $10.2~\rm{m/s}$ directly against gravity. At $45^\circ$ the rearward weight would be roughly $34.6~\rm{kN}$ giving acceleration in the flight path of a little over $13~\rm{m/s}$, but if I resolve the vertical component of that acceleration it's not enough to counteract gravity and gain altitude. So obviously I'm doing it wrong, I just don't know exactly how. Naturally I'm ignoring aerodynamic forces.

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Just use vector addition on (all) your forces to get a net force. Then you can calculate the acceleration based on it. Discounting drag, you just have thrust ($T$) and weight ($W$).

$$F_x = T_x + W_x$$ $$F_x = T \cos(\frac{\pi}{2}) + 0$$ $$F_x = 70.7 \text{kN}$$

$$F_y = T_y + W_y$$ $$F_y = T \sin(\frac{\pi}{2}) + W $$ $$F_y = 70.7 \text{kN} - 50\text{kN} = 20.7\text{kN}$$

The net vertical force ($F_y$) is positive, so it will accelerate upward.

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