# If a projectile is shot at a velocity of 7 m/s and an angle of pi/3, how far will the projectile travel before landing?

Feb 5, 2016

The projectile would travel $4.33 m$

#### Explanation:

The projectile is launched with an angle of $\frac{\pi}{3}$ which is equivalent to ${60}^{\setminus} \circ$.

The horizontal distance the projectile will travel is obtained for $y = 0$. Remember that the equation for the distance in the $y$ component is: $y = {v}_{0} \sin \left(\setminus \theta\right) t - 0.5 g {t}^{2}$.

So:

$\setminus q \quad 0 = {v}_{0} \setminus \sin \left(\setminus \theta\right) t \cdot - 0.5 g {t}^{2}$

If we solve for $t$ in the above equation:

$\setminus q \quad {v}_{0} \setminus \sin \left(\theta\right) t = 0.5 g {t}^{2}$

$\setminus q \quad {v}_{0} \setminus \sin \left(\theta\right) = 0.5 g t$

$\setminus q \quad \frac{{v}_{0} \setminus \sin \left(\theta\right)}{0.5 g} = t$

$\setminus q \quad t = \frac{2 {v}_{0} \sin \setminus \theta}{g}$

And, the equation for the distance in the $x$ component is: $x = {v}_{0} \setminus \cos \left(\theta\right) t$.

Now, if we replace $t$ in this equation, we will have:

$\setminus q \quad {x}_{\max} = {v}_{0} \setminus \cos \left(\theta\right) t$

$\setminus q \quad {x}_{\max} = {v}_{0} \setminus \cos \theta \cdot \frac{2 {v}_{0} \sin \setminus \theta}{g}$

$\setminus q \quad {x}_{\max} = \frac{2 \cdot {v}_{0}^{2} \cdot \cos \theta \cdot \sin \theta}{g}$

Remember that $2 \cos \theta \sin \theta = \sin 2 \theta$.

$\setminus q \quad {x}_{\max} = \frac{{v}_{0}^{2} \cdot \sin 2 \theta}{g}$

Finally, your maximum distance would be:

$\setminus q \quad {x}_{\max} = \frac{{v}_{0}^{2} \cdot \sin 2 \theta}{g}$

$\setminus q \quad {x}_{\max} = \frac{{7}^{2} \cdot \sin \left(2 \cdot 60\right)}{9.81} = 4.3257 m \approx 4.33 m$