# A projectile is shot at a velocity of  23 m/s and an angle of pi/6 . What is the projectile's peak height?

If $\alpha$ is the angle of projection with velocity of projection u then vertical component of this velocity is $u \sin \alpha$ and final velocity of this component will be zero
${0}^{2} = {u}^{2} {\sin}^{2} \alpha - 2 g h$
Peak height $h = \frac{{u}^{2} \times {\sin}^{2} \alpha}{2 g} = \frac{{23}^{2} \times {\sin}^{2} \left(\frac{\pi}{6}\right)}{2 \times 9.8} = 6.75 m$