# If a projectile is shot at a velocity of 2 m/s and an angle of pi/6, how far will the projectile travel before landing?

Jun 1, 2018

The distance is $= 0.35 m$

#### Explanation:

The equation describing the trajectory of the projectile in the $x - y$ plane is

$y = x \tan \theta - \frac{g {x}^{2}}{2 {u}^{2} {\cos}^{2} \theta}$

The initial velocity is $u = 2 m {s}^{-} 1$

The angle is $\theta = \left(\frac{1}{6} \pi\right) r a d$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

The distance $y = 0$

Therefore,

$x \tan \left(\frac{1}{6} \pi\right) - \frac{9.8 \cdot {x}^{2}}{2 \cdot {2}^{2} {\cos}^{2} \left(\frac{1}{6} \pi\right)} = 0$

$0.577 x - 0.0871 {x}^{2} = 0$

$x \left(0.577 - 1.633 x\right) = 0$

$x = 0$, this is the starting point

$x = \frac{0.577}{1.633} = 0.35 m$