# A projectile is shot at an angle of pi/4  and a velocity of  15 m/s. How far away will the projectile land?

Apr 15, 2018

The range is $= 23.0 m$

#### 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 $u = 15 m {s}^{-} 1$

The angle is $\theta = \left(\frac{1}{4} \pi\right) r a d$

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

The distance $y = 0$

Therefore,

$x \tan \left(\frac{1}{4} \pi\right) - \frac{9.8 \cdot {x}^{2}}{2 \cdot {15}^{2} {\cos}^{2} \left(\frac{\pi}{4}\right)} = 0$

$x - 0.044 {x}^{2} = 0$

$x \left(1 - 0.044 x\right) = 0$

$x = 0$, this is the starting point

$x = \frac{1}{0.044} = 23.0 m$

graph{x-0.044x^2 [-0.01, 14.04, -1.347, 5.673]}