# A projectile is shot at a velocity of  21 m/s and an angle of pi/8 . What is the projectile's peak height?

$= 62.1 m$
If the projectile is shot with velocity u at an angle of projection $\alpha$ with the horizontal, then it will have initial vertical component of velocity $u \sin \alpha$ and this velocity will becomes zero at maximum height H
${0}^{2} = {\left(u \cdot \sin \alpha\right)}^{2} - 2 \cdot g \cdot H$
$H = {\left(u \cdot \sin \alpha\right)}^{2} / \left(2 \cdot g\right) = {21}^{2} {\sin}^{2} \frac{\frac{\pi}{8}}{2 \cdot 9.8} = 62.1 m$