# If a projectile is shot at an angle of (5pi)/12 and at a velocity of 1 m/s, when will it reach its maximum height?

$0.04755416421469008 \setminus m$

#### Explanation:

The maximum height ${H}_{\max}$ obtained by a projectile shot at an angle $\setminus \theta = \frac{5 \setminus \pi}{12}$ with initial velocity $u = 1 \setminus \frac{m}{s}$ is given as follows

${v}^{2} = {u}^{2} + 2 a s$ (from II equation of motion)

${0}^{2} = {\left(u \setminus \sin \setminus \theta\right)}^{2} - 2 g {H}_{\max}$

${H}_{\max} = \setminus \frac{{u}^{2} \setminus \sin \setminus \theta}{2 g}$

${H}_{\max} = \setminus \frac{{1}^{2} \setminus {\sin}^{2} \left(\frac{5 \setminus \pi}{12}\right)}{2 \setminus \times 9.81}$

$= 0.04755416421469008 \setminus m$