If a projectile is shot at an angle of (pi)/8 and at a velocity of 3 m/s, when will it reach its maximum height?

Time $t = \frac{3 \cdot \sin \left(\frac{\pi}{8}\right)}{9.8} = 0.1171479895 \text{ }$second

Explanation:

For projectiles fired at angle $\theta$ with initial velocity ${v}_{0}$

$y = {v}_{0} \sin \theta \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

We are after the maximum height ${y}_{\max}$.
We differentiate $y$ with respect to $t$ then equate $\frac{\mathrm{dy}}{\mathrm{dt}} = 0$ because we maximizing height $y$

$y = {v}_{0} \sin \theta \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = {v}_{0} \sin \theta \cdot \frac{\mathrm{dt}}{\mathrm{dt}} + \frac{1}{2} \cdot g \cdot 2 \cdot t \cdot \frac{\mathrm{dt}}{\mathrm{dt}}$

$\frac{\mathrm{dy}}{\mathrm{dt}} = {v}_{0} \sin \theta \cdot 1 + \frac{1}{2} \cdot g \cdot 2 \cdot t \cdot 1$

$\frac{\mathrm{dy}}{\mathrm{dt}} = {v}_{0} \sin \theta + g \cdot t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 0$

${v}_{0} \sin \theta + g \cdot t = 0$

Solving for t

$t = \frac{- {v}_{0} \sin \theta}{g} = \frac{- 3 \cdot \sin {22.5}^{\circ}}{- 9.8} = 0.1171479895 \text{ }$second

God bless...I hope the explanation is useful.