# A projectile is shot at an angle of pi/8  and a velocity of  1 m/s. How far away will the projectile land?

Mar 13, 2016

$d = {1}^{2} / g \cdot \sin 45 = \frac{1}{\sqrt{2} g} = \frac{1}{10 \sqrt{2}} = \frac{\sqrt{2}}{20} m$

#### Explanation:

We shall use the range formula

$v = 1 \frac{m}{s} , \theta = \frac{\pi}{8} \implies 2 \theta = \frac{\pi}{4} = {45}^{\circ}$

$\textcolor{m a \ge n t a}{\text{Interesting note " 45^@ "is the optimal angle of launch}}$

$d = {1}^{2} / g \cdot \sin 45 = \frac{1}{\sqrt{2} g} = \frac{1}{10 \sqrt{2}} = \frac{\sqrt{2}}{20} m$