# If a projectile is shot at a velocity of 13 m/s and an angle of pi/6, how far will the projectile travel before landing?

Vel of projection v =$13 m {s}^{-} 1$
Angle of projection = $\theta = \frac{\pi}{6}$
Horizontal component of vel of projection =$v \cos \theta = 13 \cos \left(\frac{\pi}{6}\right) m {s}^{-} 1$
Time of flight T = $\frac{2 v \sin \theta}{g} = \frac{2 \cdot 13 \cdot \sin \left(\frac{\pi}{6}\right)}{g}$
$= v \cos \theta \cdot T$
$= {13}^{2} / g \cdot 2 \sin \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{6}\right)$
$\frac{169}{10} \cdot \frac{\sqrt{3}}{2} = 14.64 m$
[$g = 10 m {s}^{-} 2$ considered]