# If a projectile is shot at a velocity of 12 m/s and an angle of pi/3, how far will the projectile travel before landing?

Dec 18, 2016

The distance is $= 12.7 m$

#### Explanation:

We need 2 equations

$\sin 2 \theta = 2 \sin \theta \cos \theta$

$x \left(t\right) = {v}_{x} t = {u}_{0} \cos \theta t$

and

$y \left(t\right) = {v}_{y} t - \frac{1}{2} g {t}^{2}$

$y \left(t\right) = {u}_{0} \sin \theta t - \frac{1}{2} g {t}^{2}$

In our case,

$y \left(t\right) = 0$

So,

$\frac{1}{2} g {t}^{2} - {u}_{0} \sin \theta t = 0$

$t \left(\frac{1}{2} g t - {u}_{0} \sin \theta\right) = 0$

The value $t = 0$ is when the projectile is shot

$t = \frac{2 {u}_{0} \sin \theta}{g}$

Therefore,

$x \left(t\right) = {u}_{0} \cos \theta \cdot 2 {u}_{0} \sin \frac{\theta}{g}$

$= \frac{{u}_{0}^{2} \sin 2 \theta}{g}$

$= \frac{12 \cdot 12 \cdot \sin \left(\frac{2}{3} \pi\right)}{9.8}$

$= 12.7 m$

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