# A projectile is shot at a velocity of 9 m/s and an angle of pi/12 . What is the projectile's peak height?

Jan 23, 2016

$0.27679 m$

#### Explanation:

Data:-

Initial Velocity$=$ Muzzle Velocity$= {v}_{0} = 9 \frac{m}{s}$

Angle of throwing $= \theta = \frac{\pi}{12}$

Acceleration due to gravity$= g = 9.8 \frac{m}{s} ^ 2$

Height=H=??

Sol:-

We know that:

$H = \frac{{v}_{0}^{2} {\sin}^{2} \theta}{2 g}$

$\implies H = \frac{{9}^{2} {\sin}^{2} \left(\frac{\pi}{12}\right)}{2 \cdot 9.8} = \frac{81 {\left(0.2588\right)}^{2}}{19.6} = \frac{81 \cdot 0.066978}{19.6} = \frac{5.4252}{19.6} = 0.27679$

$\implies H = 0.27679 m$

Hence, the height of the projectile is $0.27679 m$