# A projectile is shot at an angle of pi/6  and a velocity of  27 m/s. How far away will the projectile land?

Feb 2, 2016

${x}_{\max} : 64 , 354 m e t e r s$

#### Explanation:

${v}_{o} :$initial velocity
$\theta :$ angle
${v}_{o} = 27 \frac{m}{s}$
$R = \frac{\pi}{6} r a d$
$t :$ traveling time from A to B

$s t e p 1 :$
$\frac{\theta}{180} = \frac{R}{\pi}$
$\frac{\theta}{180} = \frac{\pi}{6. \pi}$
$\frac{\theta}{180} = \frac{1}{6}$
$\theta = \frac{180}{6} = {30}^{o}$

$s t e p 2 :$
$g = 9.81 \frac{m}{s} ^ 2$
${v}_{x} = {v}_{o} \cos \theta$ ( no change during projectile)
${v}_{y} = {v}_{0} \sin \theta - g t$
${v}_{y} = 0$ (at maximum height)
$0 = {v}_{0} \sin \theta - g {t}_{m}$
$g {t}_{m} = {v}_{0} \sin \theta$
${t}_{m} = \frac{{v}_{0} \sin \theta}{g}$(time to maximum height)
$t = 2. {t}_{m}$(traveling time from A to B)
$t = \frac{2. {v}_{o} \sin \theta}{g}$

$s t e p 3 :$
${x}_{\max} = {v}_{x} . t$
${x}_{\max} = {v}_{o} \cos \theta . 2 \frac{{v}_{0} \sin \theta}{g}$
${x}_{\max} = {v}_{0}^{2} . 2. \sin \theta . \cos \frac{\theta}{g}$
$2. \sin \theta . \cos \theta = \sin 2 \theta$(trigonometric equation)
${x}_{\max} = {v}_{o}^{2} . \sin \frac{2 \theta}{g}$
$\theta = {30}^{o}$
$2 . \theta = {60}^{o}$
${x}_{\max} = {27}^{2} . \sin \frac{60}{9 , 18}$
$\sin 30 = 0 , 866$
${x}_{\max} = \frac{729 . 0 , 866}{9 , 86}$
${x}_{\max} = \frac{631 , 314}{9 , 86}$
${x}_{\max} = 64 , 354$ meters