# A projectile is shot at an angle of pi/3  and a velocity of  28 m/s. How far away will the projectile land?

May 20, 2017

$69.3 m$

#### Explanation:

I'll present a "shortcut" equation to find the distance a projectile will launch given an initial launch velocity and angle. The formula for maximum range $R$ of a launched projectile is given by the equation

$R = \frac{{\left({v}_{0}\right)}^{2} \sin \left(2 {\alpha}_{0}\right)}{g}$

where ${v}_{0}$ is the magnitude of the initial velocity,
${\alpha}_{0}$ is the initial launch angle, and
$g$ is the acceleration to to gravity near Earth's surface, $9.8 \frac{m}{{s}^{2}}$.

Although it doesn't necessarily matter, the equivalent angle of $\frac{\pi}{3}$ is ${60}^{o}$. We can plug in the known variables to solve for the maximum range $R$:

R = ((28m/s)^2 sin(2(60^o)))/(9.8m/(s^2)) = color(blue)(69.3m