# A projectile is shot at a velocity of 9 m/s and an angle of pi/8 . What is the projectile's peak height?

Jan 30, 2017

$\text{the answer is "h_p=0.60" m}$

#### Explanation:

$\text{the peak height of a projectile can be calculated using:}$

${h}_{p} = \frac{{v}_{i}^{2.} {\sin}^{2} \alpha}{2. g}$

"where;

${v}_{i} : \text{initial velocity ,"v_i=9" m/s}$
$\alpha = \frac{\pi}{8} \text{ , "sin alpha=0.38268343" , } {\sin}^{2} \alpha = 0.14644661$

$g = 9.81 \frac{N}{k g}$

${h}_{p} : \text{peak height}$

${h}_{p} = \frac{{9}^{2} \cdot 0.14644661}{2 \cdot 9.81}$

${h}_{p} = \frac{81 \cdot 0.14644661}{19.62}$

${h}_{p} = 0.60 \text{ m}$