# A projectile is shot at a velocity of 12 m/s and an angle of pi/12 . What is the projectile's maximum height?

##### 1 Answer
Dec 9, 2016

The height is $= 0.05 m$

#### Explanation:

At the maximum height, the vertical component of the velocity is $= 0$

The vertical component is $v = {u}_{0} \sin \theta$

${u}_{0} = 12 m {s}^{- 1}$

$\theta = \frac{\pi}{12}$

We use the equation

${v}^{2} = {u}_{0}^{2} {\sin}^{2} \theta - 2 \cdot g \cdot h$

$0 = {u}_{0}^{2} {\sin}^{2} \theta - 2 \cdot g \cdot h$

$h = \frac{{u}_{0}^{2} {\sin}^{2} \left(\frac{\pi}{12}\right)}{2 g}$

$= 144 \cdot {\sin}^{2} \frac{\frac{\pi}{12}}{2 \cdot 9.8}$

$= 0.05 m$