# A projectile is shot at an angle of pi/8  and a velocity of  65 m/s. How far away will the projectile land?

$304.539 \setminus m$

#### Explanation:

The range $R$ covered by a projectile shot at an angle $\setminus \theta = \setminus \frac{\pi}{8}$ & a velocity $u = 65 \setminus \frac{m}{s}$ is given by the following formula

$R = \setminus \textrm{h \mathmr{and} i z o n t a l v e l o c i t y} \setminus \times \setminus \textrm{t i m e o f f l i g h t}$

$R = \left(u \setminus \cos \setminus \theta\right) \left(\frac{2 u \setminus \sin \setminus \theta}{g}\right)$

$= \setminus \frac{{u}^{2} \setminus \sin 2 \setminus \theta}{g}$

$= \setminus \frac{{65}^{2} \setminus \sin 2 \left(\setminus \frac{\pi}{8}\right)}{9.81}$

$= \setminus \frac{4225 \setminus \cdot \frac{1}{\setminus} \sqrt{2}}{9.81}$

$= 304.539 \setminus m$

Jul 10, 2018

The distance is $= 304.6 m$

#### Explanation:

Another way of finding the range.

The equation of the trajectory of the projectile is

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

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

The initial velocity is $u = 65 m {s}^{-} 1$

The angle is $\theta = \frac{\pi}{8}$

Therefore,

$y = x \tan \left(\frac{\pi}{8}\right) - \left(\frac{9.8}{2 \cdot {65}^{5} \cdot {\cos}^{2} \left(\frac{\pi}{8}\right)}\right) {x}^{2}$

$y = 0.4142 x - 0.00136 {x}^{2}$

The range is when $y = 0$

$0.4142 x - 0.00136 {x}^{2} = 0$

$\left\{\begin{matrix}x = 0 \\ x = \frac{0.4142}{0.00136} = 304.6 m\end{matrix}\right.$

graph{0.4142x-0.00136x^2 [-91, 390.2, -103.2, 137.5]}