# A projectile is shot at an angle of pi/4  and a velocity of  7 m/s. How far away will the projectile land?

Mar 24, 2016

${x}_{\text{max"=4,99 " m}}$

#### Explanation:

${v}_{i} : \text{initial velocity}$
${t}_{\text{elapsed":"elapsed time for projectile}}$
${v}_{x} : \text{x component for initial velocity}$
${x}_{\text{max"=v_x*t_"elapsed}}$
${v}_{x} = {v}_{i} \cdot \cos \alpha$
${t}_{\text{elapsed}} = \frac{2 \cdot {v}_{i} \cdot \sin \alpha}{g}$

${x}_{\text{max}} = {v}_{i} \cdot \cos \alpha \cdot \frac{2 \cdot {v}_{i} \cdot \sin \alpha}{g}$
$2 \cdot \sin \alpha \cdot \cos \alpha = \sin 2 \alpha$
$\alpha = \frac{\pi}{4} \text{ "2alpha=pi/2" } \sin \left(\frac{\pi}{2}\right) = 1$
${x}_{\text{max}} = \frac{{v}_{i}^{2} \cdot \sin 2 \alpha}{g}$

${x}_{\text{max}} = \frac{{7}^{2} \cdot 1}{g}$
${x}_{\text{max}} = \frac{49}{9 , 81}$
${x}_{\text{max"=4,99 " m}}$