# A projectile is shot at an angle of pi/12  and a velocity of  24 m/s. How far away will the projectile land?

Jul 10, 2017

The distance is $= 29.4 m$

#### Explanation:

Resolving in the vertical direction ${\uparrow}^{+}$

We apply the equation of motion

$s = u t + \frac{1}{2} a {t}^{2}$

$s = 0$

The initial velocity is $u = {u}_{0} \sin \theta = 24 \sin \left(\frac{1}{12} \pi\right)$

The acceleratio is $a = - g$

Therefore,

$24 \sin \left(\frac{1}{12} \pi\right) \cdot t - \frac{1}{2} g {t}^{2} = 0$

$t = 0$ which is the starting time and the time to land is

$t = 2 \cdot 24 \sin \frac{\frac{1}{12} \pi}{g}$

$= 1.27 s$

Resolving in the horizontal direction ${\rightarrow}^{+}$

We apply the equation of motion

$d = u \cos \theta \cdot t$

$d = 24 \cos \left(\frac{1}{12} \pi\right) \cdot 1.27 = 29.4 m$