# A projectile is shot at an angle of (5pi)/12  and a velocity of  3 m/s. How far away will the projectile land?

Mar 30, 2018

The range is $= 0.46 m$

#### Explanation:

The equation of the trajectory of the projectile in the $\left(x , y\right)$ plane is

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

The initial velocity is $u = 3 m {s}^{-} 1$

The angle is $\theta = \frac{5}{12} \pi$

The acceleration due to gravity is $= 9.8 m {s}^{-} 1$

When the projectile will land when

$y = 0$

Therefore,

$x \tan \theta - \frac{g {x}^{2}}{2 {u}^{2} {\cos}^{2} \theta} = x \tan \left(\frac{5}{12} \pi\right) - \frac{9.8 {x}^{2}}{2 \cdot {3}^{2} \cdot {\cos}^{2} \left(\frac{5}{12} \pi\right)} = 0$

$x \left(3.73 - 8.13 x\right) = 0$

$x = \frac{3.73}{8.13}$

$= 0.46 m$

graph{3.73x-8.13x^2 [-6.2, 204.7, -42.2, 63.3]}