A projectile is shot at an angle of pi/6  and a velocity of  18 m/s. How far away will the projectile land?

1 Answer
Feb 3, 2016

$28.60 m$

Explanation:

we know,
$R = \frac{{v}^{2} \sin 2 \theta}{g}$

$= \frac{{18}^{2} \sin \left(\frac{\pi}{3}\right)}{9.81}$

$= 28.60 m$