# If a projectile is shot at an angle of pi/12 and at a velocity of 22 m/s, when will it reach its maximum height??

Jan 31, 2018

$0.57 s$

#### Explanation:

here,the projectile will go up due to the vertical component of its velocity, i.e $22 \sin \left(\frac{\pi}{12}\right)$ or, $5.7 \frac{m}{s}$

So,it will reach the highest point,when its upward velocity becomes zero,

so,we can apply $v = u - g t$

So,for $v = 0$, $t = \frac{5.7}{10} s$ or,$0.57 s$