The vertical component of the ball's initial velocity is
[tex]v_{0y}=\left(9\,\dfrac{\mathrm m}{\mathrm s}\right)\sin60^\circ=7.8\,\dfrac{\mathrm m}{\mathrm s}[/tex]
At any point along its trajectory, the vertical component of the ball's velocity [tex]v_y[/tex] satisfies
[tex]{v_y}^2-{v_{0y}}^2=2(-g)(y-2\,\mathrm m)[/tex]
where [tex]y[/tex] is the corresponding height of the ball while it's in the air. At its maximum height [tex]y_{\mathrm{max}}[/tex], the ball's vertical velocity is 0, so we have
[tex]-\left(7.8\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(y_{\mathrm{max}}-2\,\mathrm m)[/tex]
[tex]\implies y_{\mathrm{max}}=5.1\,\mathrm m[/tex]
