A projectile is shot at a velocity of  12 m/s and an angle of pi/6 . What is the projectile's peak height?

Jul 3, 2016

${h}_{m} = 1.83 \text{ } m$

Explanation:

$\text{The projectile's peak height can be calculated using :}$

${h}_{m} = \frac{{V}_{i}^{2} \cdot {\sin}^{2} \alpha}{2 \cdot g}$

${h}_{m} \text{ represents peak height}$

${v}_{i} : \text{initial velocity}$

${v}_{i} = 12 \text{ } \frac{m}{s}$

$\alpha = \left(\frac{\pi}{6}\right) \text{ ; "sin (pi/6)=1/2" ; } {\sin}^{2} \left(\frac{\pi}{6}\right) = \frac{1}{4}$

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

${h}_{m} = \frac{{12}^{2} \cdot \frac{1}{4}}{2 \cdot 9.81}$

h_m=36//19.62)

${h}_{m} = 1.83 \text{ } m$