Why is initial position negative? 
A student on ground throws a ball straight up, the ball leaves students hand with speed of 15m/s when hand is 2.0mm above ground How long is the ball in the air before it hits the ground?

I was able to figure out the answer to this question using $V_0 = 15\text{ m/s}$, $t_0 = 0\text{ s}$, $Y_0 = 0\text{ m}$, $Y_1 = -2\text{ m}$, $a = -9.8\ \mathrm{m/s^2}$
and using $S_f = S_i + V_i t + \frac{1}{2} a t^2$ I could find $t_1$.
My question is, why is $y_1$ negative?
I also remember learning that $a$ is always negative, at least in these kind of problems.
 A: You are probably mistaken about "hand is 2.0mm above ground".
Anyway, the convention you used here is positive values going up. Then if you call $Y_1$ the hand height, then $Y_1$ is positive. And yes, $a$ is negative here because you have to express the fact that gravity is pulling the body to the negative direction.
A: Take a look at this

This is the coordinate system we generally use to solve classical mechanics questions. In your case we would be setting the origin at the place where you launched the ball. 
Everything that points downwards, will naturally be written with "-ve" prefix. 
If $\text{Y}_1$ is the final position of the ball, you can see the ball reaches below the hand and that would be below the origin and hence measured in "negative"
Also I think you have sort of mugged up that acceleration in such questions is "negative", try to understand it. See that if everything that points up is denoted with "+ve" prefix, everything that points downwards will be denoted with "-ve" prefix. Since here the acceleration is the pull of gravitational field of earth i.e. gravity, and since it points downwards therefore it is taken as "-ve".
