# If a projectile is shot at an angle of pi/6 and at a velocity of 18 m/s, when will it reach its maximum height??

Mar 30, 2016

Time of reaching at maximum height
$t = \frac{u \sin \alpha}{g} = \frac{18 \cdot \sin \left(\frac{\pi}{6}\right)}{9.8} = 0.91 s$

Mar 30, 2016

$t \cong 0 , 92 \text{ s}$

#### Explanation:

g=9,82 " m/s^2

${v}_{i} = 18 \text{ } \frac{m}{s}$

$\alpha = \frac{\pi}{6}$

$\sin \alpha = 0 , 5$

$t = \frac{{v}_{i} \cdot \sin \alpha}{g}$

$t = \frac{18 \cdot 0 , 5}{9 , 81}$

$t = \frac{9}{9 , 81}$

$t \cong 0 , 92 \text{ s}$