# A projectile is shot at an angle of pi/6  and a velocity of  3 2 m/s. How far away will the projectile land?

Jul 12, 2016

${x}_{m} = 90.40 \text{ } m$

#### Explanation: $\text{please look over the animation carefully}$
$\text{The initial velocity of object can be split into two components}$
$\text{as vertical and horizontal. } \left({v}_{x} , {v}_{y}\right)$

${v}_{x} = {v}_{i} \cdot \cos \alpha$
${v}_{y} = {v}_{i} \cdot \sin \alpha - g \cdot t$ $\text{we can find the time elapsed using the component } {v}_{y}$

$\text{we can write "v_y=0 " for the maximum height.}$

${v}_{y} = {v}_{i} \cdot \sin \alpha - g \cdot t = 0$

${v}_{i} \cdot \sin \alpha = g \cdot t$

$t = \frac{{v}_{i} \cdot \sin \alpha}{g}$

${t}_{f} : \text{represents the flaying time}$

${t}_{f} = 2 \cdot t = \frac{2 \cdot {v}_{i} \cdot \sin \alpha}{g}$ ${x}_{m} = {v}_{x} \cdot {v}_{f}$

${x}_{m} = \frac{{v}_{i} \cdot \cos \alpha \cdot 2 \cdot {v}_{i} \cdot \sin \alpha}{g}$

${x}_{m} = \frac{{v}_{i}^{2} \cdot 2 \cdot \sin \alpha \cdot \cos \alpha}{g}$

$2 \cdot \sin \alpha \cdot \cos \alpha = \sin 2 \alpha$

${x}_{m} = \frac{{v}_{i}^{2} \cdot \sin 2 \alpha}{g}$

$\text{where:}$
${v}_{i} = 32 \text{ } \frac{m}{s}$
$2 \cdot \alpha = \frac{\pi}{6} \cdot 2 = \frac{\pi}{3}$
$g = 9.81 \text{ } \frac{m}{s} ^ 2$

${x}_{m} = \frac{{32}^{2} \cdot 0.866}{9.81}$

${x}_{m} = 90.40 \text{ } m$