# A projectile is shot at an angle of pi/6  and a velocity of  3 9 m/s. How far away will the projectile land?

Mar 18, 2018

Here the required distance is nothing but the range of the projectile motion,which 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, $u = 39 m {s}^{-} 1 , \theta = \frac{\pi}{6}$

So,putting the given values we get,

$R = 134.4 m$

Mar 18, 2018

$\text{134.4 m}$

#### Explanation:

Range ($\text{R}$) of a projectile is given as

$\text{R" = ("u"^2 sin(2theta))/"g}$

$\text{R" = ("(39 m/s)"^2 × sin(2 xx π/6))/("9.8 m/s"^2) = "134.4 m}$