# A projectile is shot at a velocity of  18 m/s and an angle of pi/6 . What is the projectile's peak height?

Mar 15, 2018

The peak height is $= 4.13 m$

#### Explanation:

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

The initial velocity is ${u}_{0} = \left(18\right) \cdot \sin \left(\frac{1}{6} \pi\right) m {s}^{-} 1$

Applying the equation of motion

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

At the greatest height, $v = 0 m {s}^{-} 1$

The acceleration due to gravity is $a = - g = - 9.8 m {s}^{-} 2$

Therefore,

The greatest height is ${h}_{y} = s = {\left(0 - \left(18\right) \sin \left(\frac{1}{6} \pi\right)\right)}^{2} / \left(- 2 g\right)$

${h}_{y} = {\left(18 \sin \left(\frac{1}{6} \pi\right)\right)}^{2} / \left(2 \cdot 9.8\right) = 4.13 m$