How high will it go?

$y=20\ \mathrm{m}$

We are given $v_0$, $a = –10\ \mathrm{\displaystyle \frac{m}{\ s^2}}$ is implied, and we are asked for $y$. Now, neither $t$ nor $v$ is expressly given; however, we know the vertical velocity at the top is $0$ (otherwise the baseball would still rise). Consequently, we use Big Five equation #5.

$$v^2 = {v_0}^2+2a(y-y_0)$$

$$−2ay ={v_0}^2$$

We set $y_0 = 0$ and we know that $v = 0$, so that leaves us with:

$$y=-\frac{{v_0}^2}{2a}$$

$$y=-\frac{\left(20\ \mathrm{\displaystyle \frac{m}{s}}\right)^2}{2\left(-10\ \mathrm{\displaystyle \frac{m}{\ s^2}}\right)}$$

$$y=20\ \mathrm{m}$$