A projectile is shot at an angle of pi/8  and a velocity of  7 m/s. How far away will the projectile land?

Aug 8, 2018

The projectile wiil land at $= 3.54 m$

Explanation:

The trajectory of a projectile is given by the equation

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

Here,

The initial velocity is $u = 7 m {s}^{-} 1$

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

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

Therefore,

$y = x \tan \left(\frac{\pi}{8}\right) - \left(\frac{9.8}{2 \cdot {7}^{2} {\cos}^{2} \left(\frac{\pi}{8}\right)}\right) {x}^{2}$

$y = 0.414 x - 0.117 {x}^{2}$

The distance travelled horizontally is when $y = 0$

$0 = 0.414 x - 0.117 {x}^{2}$

$x \left(0.414 - 0.117 x\right) = 0$

$x = 0$, this is the initial conditions

$x = \frac{0.414}{0.117} = 3.54 m$

graph{0.414x-0.117x^2 [-0.18, 2.858, -0.269, 1.249]}