A projectile is shot at an angle of pi/12  and a velocity of 45 m/s. How far away will the projectile land?

Jan 11, 2016

Range$= 103.316 m$

Explanation:

The range of a projectile is given by
$R = \frac{{v}_{0}^{2} \cdot S \in 2 \theta}{g}$

Where $R$ is the range, ${v}_{0}$ is the initial or muzzle velocity, $\theta$ is the throwing angle and $g$ is the acceleration due to gravity.
Here ${v}_{0} = 45 \frac{m}{s}$, $\theta = \frac{\pi}{12}$, $g = 9.8 \frac{m}{s} ^ 2$ and R=??

$\implies R = \frac{{45}^{2} \cdot S \in \left(2 \cdot \frac{\pi}{12}\right)}{9.8} = \frac{2025 \cdot S \in \left(\frac{\pi}{6}\right)}{9.8} = \frac{\left(2025\right) 0.5}{9.8} = 103.316 m$
i
$\implies R = 103.316 m$