# A projectile is shot at an angle of pi/6  and a velocity of  3 m/s. How far away will the projectile land?

Range $x = 0.795329452 \text{ }$meters

#### Explanation:

Given $\theta = \frac{\pi}{6}$
initial velocity ${v}_{0} = 3 \text{ }$m/sec

Compute the total time t it will be on the air

$y = {v}_{0} \cdot \sin \theta \cdot t + \frac{1}{2} \cdot g \cdot {t}^{2}$

$0 = 3 \cdot \sin \left(\frac{\pi}{6}\right) \cdot t + \frac{1}{2} \left(- 9.8\right) \cdot {t}^{2}$

$0 = 3 \cdot \left(0.5\right) \cdot t - 4.9 \cdot {t}^{2}$

$0 = 1.5 \cdot t - 4.9 \cdot {t}^{2}$

$0 = t \left(1.5 - 4.9 t\right)$

${t}_{1} = 0$ and ${t}_{2} = \frac{1.5}{4.9}$

Let $x =$range

We will use ${t}_{2} = \frac{1.5}{4.9}$

$x = {v}_{0} \cos \theta \cdot t$

$x = 3 \cdot \cos \left(\frac{\pi}{6}\right) \left(\frac{1.5}{4.9}\right)$

$x = 0.795329452 \text{ }$meters

God bless....I hope the explanation is useful.