# A projectile is shot at an angle of (5pi)/12  and a velocity of  72 m/s. How far away will the projectile land?

Aug 19, 2017

see below

#### Explanation:

Apply the formula for horizontal range with $\theta = \frac{5 \pi}{12}$ and ${v}_{0} = 72 \frac{m}{s}$

Horizontal range, $R = \frac{{v}_{0}^{2} \sin \left(2 \theta\right)}{g}$

$\implies R = \frac{{\left(72\right)}^{2} \cdot \sin \left(2 \cdot \frac{5 \pi}{12}\right)}{9.8} = 264.49$ $m$

Hence, it will land at a distance of $264.49$ $m$ from the point of projection.