# If a projectile is shot at a velocity of 4 m/s and an angle of pi/6, how far will the projectile travel before landing?

$1.412 \setminus m$

#### Explanation:

The horizontal distance/ range $R$ traveled by a projectile shot with an initial velocity u=4\ \text{m/s at an angle $\setminus \theta = \setminus \frac{\pi}{6}$ with the horizontal is given by following formula

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

$= \setminus \frac{{4}^{2} \setminus \sin \left(\setminus \frac{\pi}{3}\right)}{9.81}$

$= 1.412 \setminus m$

Jul 28, 2018

The distance is $= 1.41 m$

#### 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 $u = 4 m {s}^{-} 1$

The angle is $\theta = \frac{\pi}{6}$

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

Therefore,

$y = x \tan \left(\frac{\pi}{6}\right) - \left(\frac{9.8}{2 \cdot {4}^{2} {\cos}^{2} \left(\frac{\pi}{6}\right)}\right) {x}^{2}$

$y = 0.577 x - 0.408 {x}^{2}$

The distance travelled horizontally is when $y = 0$

$0 = 0.577 x - 0.408 {x}^{2}$

$x \left(0.577 - 0.408 x\right) = 0$

$x = 0$, this is the initial conditions

$x = \frac{0.577}{0.408} = 1.41 m$

graph{0.577x-0.408x^2 [-0.18, 2.858, -0.269, 1.249]}