If a projectile is shot at an angle of (pi)/8 and at a velocity of 35 m/s, when will it reach its maximum height?

First find the vertical velocity ${v}_{y} = v \sin \theta$
Where $\theta = \frac{\pi}{8} \mathmr{and} \frac{45}{2}$
${v}_{f}^{2} = {v}_{i}^{2} + 2 a d$
Where v_f=0; a=g=9.81m/s^2
$d = \sqrt{{v}_{y}^{2} / \left(2 g\right)}$
=sqrt(((v^2sin^2theta)/(2g))