Velocity-time-distance problem In my book the formula for the $y$-component of velocity during the upward projectile motion is given:
$$V_y=V_{iy}-gt$$
and next to it the formula for $y$-component of velocity during the downward projectile motion is given, differed only by a conjugate,
$$V_y=V_{iy}+gt$$
I think it must have been:
$$V_y=gt$$
since $V_{iy}=0$ at the maximum height of projectile. I took time arbitrarily same for both conditions i.e. $t$.
Am I right?
 A: It's customary to use small $v$ for velocity (but not mandatory).
Notations like $v_y$ and $v_{iy}$ make little sense without defining them. Here it implies the velocity component along the $y$-axis but without specifying this axis nothing makes much sense either.
So I'll define the $y$-axis as a vertical axis, pointing upwards. This sense is important too, as we'll define the scalar of any vector pointing in that direction as positive and the scalar of any vector pointing in the opposite direction as negative:

Having cleared this up, the expression:
$$v_y=v_{iy}-gt,$$
now makes perfect sense because $\vec{g}$ points downwards. The index iy of course refers to initial.
It's also the only expression (for speed) we need. Assuming $\vec{v_{iy}}$ pointed upward then $v_y$ is positive as long as:
$$v_{iy}>gt,$$
after which the object starts falling back to Earth and $v_y$ becomes negative in accordance with the convention agreed upon higher up.
Note also that if $\vec{v_{iy}}$ was pointing downwards (an object thrown downwards from a 
building for instance) the expression for $v_y$ still returns the correct result.
The expression can be further generalised as:
$$v_y=v_{iy}+at.$$
For numerical computations, the correct signs then have to be assigned to the values of $v_{iy}$ and $a$, in accordance with the directions of the vectors $\vec{v_{iy}}$ and $\vec{a}$.
