# A projectile is shot from the ground at an angle of pi/6  and a speed of 15 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Mar 8, 2016

$t = \frac{15 \cdot 0 , 5}{9 , 81} = 0 , 76 \text{ } s$
$x \cong 9 , 87 m$
$y = {h}_{m} \cong 2 , 87 m$
${v}_{y} = 0 \text{ at the point of D}$
${v}_{x} = 12 , 99 \frac{m}{s}$

#### Explanation:

$\alpha : \frac{\pi}{6} = {30}^{o} \text{ } {v}_{i} = 15 \frac{m}{s}$
$t = {v}_{i} \cdot \sin \frac{\alpha}{g} \text{ 'elapsed time to the maximum height'}$
$t = \frac{15 \cdot 0 , 5}{9 , 81} = 0 , 76 \text{ } s$
$y = {h}_{m} = {v}_{i} \cdot t \cdot \sin \alpha - \frac{1}{2} \cdot g \cdot {t}^{2}$
$y = {h}_{m} = 15 \cdot 0 , 76 \cdot 0 , 5 - \frac{1}{2} \cdot 9 , 81 \cdot {\left(0 , 76\right)}^{2}$
$y = {h}_{m} = 5 , 7 - 2 , 83$
$y = {h}_{m} \cong 2 , 87 m$
$x = {v}_{i} \cdot t \cdot \cos \alpha$
$x = 15 \cdot 0 , 76 \cdot 0 , 866$
$x \cong 9 , 87 m$
${v}_{y} = 0 \text{ at the point of D}$
${v}_{x} = {v}_{i} \cdot \cos \alpha$
${v}_{x} = 15 \cdot 0 , 866$
${v}_{x} = 12 , 99 \frac{m}{s} \text{ The x component of velocity is not change}$