Answer
With ground level chosen so $y_{0}=0,$ this equation gives the result
$$
y_{\max }=\frac{1}{2} g t_{1}^{2}=\frac{1}{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(3.0 \mathrm{s})^{2}=44.1 \mathrm{m}
$$
Work Step by Step
With ground level chosen so $y_{0}=0,$ this equation gives the result
$$
y_{\max }=\frac{1}{2} g t_{1}^{2}=\frac{1}{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)(3.0 \mathrm{s})^{2}=44.1 \mathrm{m}
$$