# If a projectile is shot at a velocity of 45 m/s and an angle of pi/6, how far will the projectile travel before landing?

Mar 17, 2018

Range of projectile motion is given by the formula $R = \frac{{u}^{2} \sin 2 \theta}{g}$ where,$u$ is the velocity of projection and $\theta$ is the angle of projection.

Given, $v = 45 m {s}^{-} 1 , \theta = \frac{\pi}{6}$

So, $R = \frac{{45}^{2} \sin \left(\frac{\pi}{3}\right)}{9.8} = 178.95 m$

This is the displacement of the projectile horizontally.

Vertical displacement is zero,as it returned to the level of projection.

Mar 17, 2018

The projectile will travel $= 178.94 m$

#### Explanation:

The equation of the trajectory of the projectile in the $\left(x , y\right)$ plane is

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

The initial velocity is $u = 45 m {s}^{-} 1$

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

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

When the projectile will land when

$y = 0$

Therefore,

$x \tan \theta - \frac{g {x}^{2}}{2 {u}^{2} {\cos}^{2} \theta} = x \tan \left(\frac{\pi}{6}\right) - \frac{9.8 {x}^{2}}{2 \cdot {45}^{2} \cdot {\cos}^{2} \left(\frac{\pi}{6}\right)} = 0$

$x \left(0.577 - 0.0032 x\right) = 0$

$x = \frac{0.577}{0.0032}$

$= 178.94 m$

graph{0.577x-0.0032x^2 [-6.2, 204.7, -42.2, 63.3]}