# If a projectile is shot at a velocity of 18 m/s and an angle of pi/12, how far will the projectile travel before landing?

Feb 6, 2017

The distance is $= 16.53 m$

#### Explanation:

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

$u = 18 \sin \left(\frac{\pi}{12}\right)$

$v = 0$

$a = - g$

Time of flight to reach greatest height, $t$

We apply,

$v = u + a t$

$0 = 18 \sin \left(\frac{\pi}{12}\right) - g t$

$t = \frac{18}{g} \sin \left(\frac{\pi}{12}\right)$

The time to land is $= 2 t = \frac{36}{g} \sin \left(\frac{\pi}{12}\right)$

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

$u = 18 \cos \left(\frac{\pi}{12}\right)$

Distance $s = u \cdot 2 t = 18 \cos \left(\frac{\pi}{12}\right) \cdot \frac{36}{g} \sin \left(\frac{\pi}{12}\right)$

$= {18}^{2} / g \cdot \sin \left(\frac{\pi}{6}\right) = 16.53 m$