# A projectile is shot at an angle of pi/3  and a velocity of  8 m/s. How far away will the projectile land?

${x}_{\max} = x = 5.655676106 \text{ }$meters

#### Explanation:

Solve for the time first

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

Assuming level ground then $y = 0$. Going up then going down then $y = 0$

$0 = 8 \sin \left(\frac{\pi}{3}\right) t + \frac{1}{2} \cdot \left(- 9.8\right) \cdot {t}^{2}$

$8 \cdot \frac{\sqrt{3}}{2} \cdot t - 4.9 \cdot {t}^{2} = 0$

$t \left(4 \sqrt{3} - 4.9 \cdot t\right) = 0$

${t}_{1} = 0 \text{ }$seconds and ${t}_{2} = \frac{4 \sqrt{3}}{4.9} = 1.413919027 \text{ }$seconds

We can solve for the range $x$ now

$x = {v}_{0} \cos \theta \cdot {t}_{2}$

$x = 8 \cdot \cos \left(\frac{\pi}{3}\right) \cdot 1.413919027$

$x = 5.655676106 \text{ }$meters

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