# A projectile is shot at an angle of pi/12  and a velocity of  7 m/s. How far away will the projectile land?

Apr 23, 2017

The distance is $= 2.5 m$

#### Explanation:

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

initial velocity is ${u}_{y} = v \sin \theta = 7 \cdot \sin \left(\frac{1}{12} \pi\right)$

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

$0 = 7 \sin \left(\frac{1}{12} \pi\right) - g \cdot t$

$t = \frac{7}{g} \cdot \sin \left(\frac{1}{12} \pi\right)$

$= 0.185 s$

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

To find the distance where the projectile will land, we apply the equation of motion

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

$= 7 \cos \left(\frac{1}{12} \pi\right) \cdot 0.185 \cdot 2$

$= 2.5 m$