If a projectile is shot at a velocity of 12 m/s and an angle of pi/12, how far will the projectile travel before landing?

Apr 7, 2018

The range is $= 7.34 m$

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 $u = 12 m {s}^{-} 1$

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,

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

$0.268 x - 0.0365 {x}^{2} = 0$

$x \left(0.268 - 0.0365 x\right) = 0$

$x = 0$, this is the starting point

$x = \frac{0.268}{0.0365} = 7.34 m$

graph{0.268x-0.0365x^2 [-0.01, 14.04, -1.347, 5.673]}