# A projectile is shot at a velocity of  16 m/s and an angle of pi/6 . What is the projectile's peak height?

Dec 27, 2017

The peak height is $= 3.27 m$

#### Explanation:

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

The initial velocity is $u = 16 \sin \left(\frac{1}{6} \pi\right) m {s}^{-} 1$

Let the peak height be $= h m$

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

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

Applying the equation of motion

${v}^{2} = {u}^{2} + 2 a s = {u}^{2} - 2 g h$

$h = \frac{{u}^{2} - {v}^{2}}{2 g} = \frac{{\left(16 \sin \left(\frac{1}{6} \pi\right)\right)}^{2} - 0}{2 g} = 3.27 m$