# A projectile is shot at a velocity of  4 m/s and an angle of pi/6 . What is the projectile's peak height?

Mar 29, 2017

The peak height is $= 0.204 m$

#### Explanation:

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

$u = {u}_{0} \sin \theta = 4 \sin \left(\frac{1}{6} \pi\right)$

$a = - g$

$v = 0$ at the maximum height

We apply the equation

${v}^{2} = {u}^{2} + 2 a h$

$0 = 16 {\sin}^{2} \left(\frac{\pi}{6}\right) - 2 g h$

$h = \frac{1}{2 g} \cdot 16 {\sin}^{2} \left(\frac{\pi}{6}\right)$

$= 0.204 m$