Answer
0.459 m
Work Step by Step
The maximum height of a projectile is given by
$h_{m}=\frac{(v_{0}\sin\theta_{0})^{2}}{2g}$
Initial velocity $v_{0}=3.00\,m/s$
The projection angle(made with x-axis) $\theta_{0}=90^{\circ}$
Therefore, $h_{m}=\frac{(3.00\,m/s\times\sin90^{\circ})^{2}}{2\times9.8\,m/s^{2}}=0.459\,m$
