Answer: 34.5

When the ball is thrown with velocity $v$, the maximum height attained is given by $y_{max}= \large\frac{v_y^2}{2g}$

Here $v_y$ is the $y$ component of the initial velocity $ = v_0\sin \theta = 30 \times \sin 60^{\circ} = 26\;m/s$

$\Rightarrow y_{max} = \large\frac{26^2}{2 \times 9.8}$$ = 34.5m$