# A projectile is shot at a velocity of  7 m/s and an angle of pi/8 . What is the projectile's peak height?

Jun 2, 2016

${h}_{\text{peak"=0.37" }} m$

#### Explanation:

$\text{the peak height of a projectile can be calculated using the formula :}$
${h}_{\text{peak}} = \frac{{v}_{i}^{2} \cdot {\sin}^{2} \alpha}{2 \cdot g}$

${v}_{i} = 7 \text{ } \frac{m}{s}$

$\alpha = \frac{\pi}{8}$

$\sin \left(\frac{\pi}{8}\right) = 0.383$

$g = 9.81 \text{ } \frac{m}{s} ^ 2$

${h}_{\text{peak}} = \frac{{7}^{2} \cdot {\left(0.383\right)}^{2}}{\left(2 \cdot 9.81\right)}$

${h}_{\text{peak}} = \frac{49 \cdot 0.147}{19.62}$

${h}_{\text{peak"=0.37" }} m$