# If a projectile is shot at an angle of (2pi)/3 and at a velocity of 26 m/s, when will it reach its maximum height?

Apr 4, 2016

${t}_{m} = 2 , 29 \text{ s}$

#### Explanation:

$\text{given data:}$
$\text{initial velocity :} {v}_{i} = 26 \frac{m}{s}$
$\text{angle :} \alpha = \frac{2 \pi}{3}$
$\sin \frac{2 \pi}{3} = 0 , 866$
$\text{elapsed time to reach its maximum height :} {t}_{m} = \frac{{v}_{i} \cdot \sin \alpha}{g}$
$g = 9 , 81 \frac{m}{s} ^ 2$

${t}_{m} = \frac{26 \cdot 0 , 866}{9 , 81}$

${t}_{m} = 2 , 29 \text{ s}$