# If a projectile is shot at a velocity of 4 m/s and an angle of pi/3, how far will the projectile travel before landing?

May 7, 2017

The distance is $= 1.41 m$

#### Explanation:

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

initial velocity is ${u}_{y} = v \sin \theta = 4 \cdot \sin \left(\frac{1}{3} \pi\right)$

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

$0 = 4 \sin \left(\frac{1}{3} \pi\right) - g \cdot t$

$t = \frac{4}{g} \cdot \sin \left(\frac{1}{3} \pi\right)$

$= 0.352 s$

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

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

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

$= 4 \cos \left(\frac{1}{3} \pi\right) \cdot 0.352 \cdot 2$

$= 1.41 m$