# A projectile is shot at an angle of pi/12  and a velocity of 3 6 m/s. How far away will the projectile land?

Jan 24, 2016

Data:-

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

Initial Velocit$+$ Muzzle Velocity$= {v}_{0} = 36 \frac{m}{s}$

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

Range=R=??

Sol:-

We know that:

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

$\implies R = \frac{{36}^{2} \sin \left(2 \cdot \frac{\pi}{12}\right)}{9.8} = \frac{1296 \sin \left(\frac{\pi}{6}\right)}{9.8} = \frac{1296 \cdot 0.5}{9.8} = \frac{648}{9.8} = 66.1224 m$

$\implies R = 66.1224 m$