As the title states, why is acceleration due to gravity always (-). Say you assign "up" as the positive direction. If an projectile is thrown at a 24 degree angle above the horizontal, I get that acceleration due to gravity before the vertex is negative. However, why is it not positive after the vertex? If acceleration due to gravity is negative and we assign downwards as negative, wouldn't that make acceleration positive?
What I think is that acceleration due to gravity is always towards the ground. Even if a projectile is going downwards, and we assign downwards as (-), the acceleration due to gravity is still (-), because the object still accelerating downwards. Despite it going upwards or downwards, the net acceleration of the object is downwards.
Best Answer
It seems your misunderstanding is in understanding the concept of frame of reference. When we do calculations in physics we do this with respect to a coordinate system/frame of reference which you can chose freely (but preferably conveniently). All quantities such as position, velocity, acceleration are measured/calculated with respect to this coordinate system.
Your questions suggest that you want to consider acceleration with respect to the direction of the velocity (which does change direction itself). Your proposal is like starting with a coordinate system and once the object reaches the vertex you flip/mirror/reverse the axes of your coordinate system.
Taking your example of throwing/shooting a projectile up vertically. Let's chose the coordinates such that positive $x$ direction is up. Then, by definition the velocity at any time is $$v=\frac{dx}{dt}$$ and the acceleration is $$a=\frac{dv}{dt}=\frac{d^2x}{dt^2}$$
Before reaching the vertex, going up
Obviously $v>0$ since the position $x$ is increasing ($dx>0$). Since the projectile is decelerating $dv<0$ and therefore $a<0$.
After reaching the vertex, falling down
$v<0$, the projectile is going down ($dx<0$). The projectile is accelerating, i.e. $|v(t+dt)|-|v(t)| > 0$, but since the velocity is negative this can be written as $v(t)-v(t+dt)>0$. Therefor $a=\lim_{dt\to 0}(v(t+dt)-v(t))/dt<0$
Of course you can do the same reasoning in a different coordinate system where the $x$-axis is pointing down.