A ball is thrown straight up at 12 m/s. Take g = 10 m/s².

Without air, everything falls with the same acceleration $g$ ($9.8\ \text{m/s}^2$, or $10\ \text{m/s}^2$ when the problem says so). Thrown up at $u$: it stops at the top after $\dfrac{u}{g}$, at a height $\dfrac{u^2}{2g}$, and takes as long to come down.

For steady acceleration: $v = u + at$, $s = ut + \tfrac12 at^2$, $v^2 = u^2 + 2as$, and $s = \tfrac12(u + v)t$.