# A projectile is shot at an angle of pi/12  and a velocity of 3 2 m/s. How far away will the projectile land?

Apr 26, 2018

The distance is $= 52.25 m$

#### 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 $u = 32 m {s}^{-} 1$

The angle is $\theta = \left(\frac{1}{12} \pi\right) r a d$

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

The distance $y = 0$

Therefore,

$x \tan \left(\frac{1}{12} \pi\right) - \frac{9.8 \cdot {x}^{2}}{2 \cdot {32}^{2} {\cos}^{2} \left(\frac{1}{12} \pi\right)} = 0$

$0.268 x - 0.0051 {x}^{2} = 0$

$x \left(0.268 - 0.0051 x\right) = 0$

$x = 0$, this is the starting point

$x = \frac{0.268}{0.0051} = 52.25 m$

graph{0.268x-0.0051x^2 [-5, 68.1, -15.73, 20.8]}