# If a projectile is shot at an angle of pi/12 and at a velocity of 2 m/s, when will it reach its maximum height??

Jul 19, 2017

The time to reach the greatest height is $= 0.053 s$

#### Explanation:

Resolving in the vertical direction ${\uparrow}^{+}$

The initial velocity is ${u}_{y} = v \sin \theta = 2 \cdot \sin \left(\frac{1}{12} \pi\right)$

The Acceleration is $a = - g$

At the maximum height, $v = 0$

We apply the equation of motion

$v = u + a t$

to calculate the time to reach the greatest height

$0 = 2 \sin \left(\frac{1}{12} \pi\right) - g \cdot t$

$t = \frac{2}{g} \cdot \sin \left(\frac{1}{12} \pi\right)$

$= 0.053 s$