# A projectile is shot at a velocity of  21 m/s and an angle of pi/6 . What is the projectile's peak height?

Mar 10, 2016

${h}_{m} = 1 , 124 m$

#### Explanation:

${v}_{i} = 21 \frac{m}{s} \text{ "alpha=pi/6" } \sin \left(\frac{\pi}{6}\right) = 0 , 5$
${h}_{m} = \frac{{v}_{i}^{2} \cdot {\sin}^{2} \alpha}{2 g}$
${h}_{m} = \frac{{21}^{2} \cdot {\left(0 , 5\right)}^{2}}{2 \cdot 9 , 81}$
${h}_{m} = \frac{22 , 05}{19 , 62}$
${h}_{m} = 1 , 124 m$