Say you have the classic problem of a projectile going at 10m/s at 15 degrees off a plane already inclined at 30 degrees (so the projectile's angle with respect to the earth is 45 degrees), and you are asked for the time the projectile was in the air.
Imagine one were to tilt their head 30 degrees, so that the "hill" is no longer inclined but is now flat earth. Regardless of how you tilt your head, earth's 9.8 m/s/s acceleration will be pointed 30 degrees skew from perpendicular to the hill. However it stands to reason that you can "convert" earth's acceleration to match the hill's normal vector.
In the new perspective, gravity's "downward force" (perpendicular to the hill) is now 8.49 m/s/s, and now there's a horizontal acceleration pulling left 4.9 m/s/s. You also have to convert your initial velocity of 10 m/s with a 15 degree incline giving a "downward" velocity of 2.59 m/s.
Now you know acceleration (8.49 m/s/s), velocity (2.59 m/s), and your change in "height"/distance from the hill's surface (0 m). Since you need time in the air, you can use the kinematic equation:
$$\Delta x = V_iT + \frac 1 2 aT^2$$
Solving this equation yields the wrong answer. I don't understand why this approach doesn't work considering that all I'm trying to find is air-time, not displacement. What I'd expect to be wrong is the conversion of gravitational acceleration from the tilt, but I can't find the problem there. I'm not asking how to best work the problem, I'm asking why this method doesn't work.