# If a projectile is shot at a velocity of 2 m/s and an angle of pi/6, how far will the projectile travel before landing?

Jan 8, 2018

$0.35$ meters

#### Explanation:

The vertical component of velocity at firing will be

$2 \sin \left(\frac{\pi}{6}\right) = 2 \left(\frac{1}{2}\right) = 1 \frac{m}{s}$

The horizontal component will be

$2 \cos \left(\frac{\pi}{6}\right) = 2 \frac{\sqrt{3}}{2} = \sqrt{3} \frac{m}{s}$

Now we use the equation $y = {y}_{0} + {v}_{y 0} t + \frac{1}{2} {a}_{y} {t}^{2}$

We know that the only acceleration in projectile motion (neglecting air resistance and friction) is due to gravity. Also, we know that $y = {y}_{0} = 0$, letting $0$ be the ground.

$0 = {v}_{y 0} t + \frac{1}{2} {a}_{y} {t}^{2}$

$0 = t + \frac{1}{2} \left(- 9.8\right) {t}^{2}$

$0 = - 4.9 {t}^{2} + t$

$0 = t \left(- 4.9 t + 1\right)$

$t = 0 \mathmr{and} 0.204$

So the projectile will hit the ground after $0.204$ seconds. Now we use the same kinematic equation although this time in $x$.

$x = 0 + \sqrt{3} \left(0.204\right) + \frac{1}{2} {a}_{x} {t}^{2}$

There will be no acceleration in this case (there never is in the horizontal direction).

Therefore, $x$ is simply

$x = \sqrt{3} \left(0.204\right)$

$0.35$ m

$\therefore$The projectile will land $0.35$ meters away from the launch point.

Hopefully this helps!