# If a projectile is shot at a velocity of 15 m/s and an angle of pi/12, how far will the projectile travel before landing?

Jun 12, 2017

The distance is $= 11.48 m$

#### Explanation:

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

initial velocity is ${u}_{y} = v \sin \theta = 15 \cdot \sin \left(\frac{1}{12} \pi\right)$

Acceleration is $a = - g$

At the maximum height, $v = 0$

We apply the equation of motion

$v = u + a t$

to calculate the time to reach the greatest height

$0 = 15 \sin \left(\frac{1}{12} \pi\right) - g \cdot t$

$t = \frac{15}{g} \cdot \sin \left(\frac{1}{12} \pi\right)$

$= 0.396 s$

To find the horizontal distance, we apply the equation of motion

$s = 2 {u}_{x} \cdot t$

$= 2 \cdot 15 \cos \left(\frac{1}{12} \pi\right) \cdot 0.396$

$= 11.48 m$