Confusion about $-g$ in formula Observe a very simple case:
$$V=V_{0}+at\quad(*)$$
In $(*)$, the $+$ sign has nothing to do with direction, because it comes from:
$$a=\frac{V-V_{0}}{t-0}=\frac{\Delta V}{\Delta t},$$
and the $-$ in this definition is to compute the change, also not about direction.
But, I've found that:
$$V_{y}=V_{0y}-gt,$$
which as a formula in some books.
And in this context I guess the $-$ sign is now telling me it is opposite to the direction of the assumption the author made.
Is the logic above correct?
Edit: another question:
Is there any formula of theory that the $+,-$ signs has to do with the direction?
 A: The equation $v = v_o +at$ comes from the definition of a constant acceleration  $a = \dfrac{v-v_o}{t}$.
To use it you need first to choose a positive direction.  
As an example suppose that a ball is thrown vertically upwards at $30 \;ms^{-1}$ and you are asked its velocity and position after 2 seconds with $g=10\;ms^{-2}$.
If you assume up as the positive direction, then
$$
\left\{ 
\begin{array}{l}
v_0 = +30 \frac{\text{m}}{\text{s}} \\
v = \; ? \\ 
a = -g = -10 \frac{\text{m}}{\text{s}^2} \\
s = \; ? \\
t = 2 \text{s}
\end{array}
\right. 
$$
$$v = v_0+at \implies v = 30+(-10)2 = +10\frac{\text{m}}{\text{s}} \ \ \ \text{(upward)}$$
$$s = \left(\frac{v+v_0}{2}\right)t \implies s = \left(\frac{30+10}{2}\right)\cdot 2 = +40 \text{m} \ \ \ \text{(upward)}$$
And if you assume down as the positive direction, then
$$
\left\{ 
\begin{array}{l}
v_0 = -30 \frac{\text{m}}{\text{s}} \\ 
v = \; ? \\ 
a = g = +10 \frac{\text{m}}{\text{s}^2} \\
s = \; ? \\
t = 2 \text{s}
\end{array}
\right. 
$$
$$v = v_0+at \implies v = -30+(+10)2 = -10\frac{\text{m}}{\text{s}} \ \ \ \text{(upward)}$$
$$s = \left(\frac{v+v_0}{2}\right)t \implies s = \left(\frac{-30+(-10)}{2}\right)\cdot 2 = -40\text{m} \ \ \ \text{(upward)}$$
A: Yes, the - sign here is based on a decision to define the direction of the velocity as up whereas the acceleration due to gravity is downwards.
A: You can choose whatever sign (+ or -) for the acceleration as long as you respect it, but the equation should always be $v=v_0+at$, even when it is $v=v_0+gt$, because $g$ is already negative by convention.
