Recall that
[tex]{v_y}^2-{v_{0y}}^2=2a_y(y-y_0)[/tex]
At its highest point, the ball will have 0 vertical velocity, so that
[tex]-\left(21.3\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.80\,\dfrac{\mathrm m}{\mathrm s^2}\right)y_{\mathrm{max}}[/tex]
[tex]\implies y_{\mathrm{max}}=23.1\,\mathrm m[/tex]
