# A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 5/8 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?

##### 1 Answer
Apr 3, 2016

${v}_{x} = 0 , 16 \text{ m/s}$
${v}_{y} = 0 , 60 \text{ m/s}$
$\text{x-component of velocity at any time:} {v}_{x} = {v}_{i} \cdot \cos \alpha$
$\text{y-component of velocity at any time:} {v}_{y} = {v}_{i} \cdot t \cdot \sin \alpha - g \cdot t$
${x}_{m} = 0 , 02 \text{ } m$
$t = 0 , 06 \text{ s}$

#### Explanation:

$\text{x-component of the initial velocity : } {v}_{x} = {v}_{\cdot} \cos \alpha$

${v}_{x} = \frac{5}{8} \cdot \cos \left(\frac{5 \pi}{12}\right)$

${v}_{x} = 0 , 16 \text{ m/s}$

$\text{y-component of the initial velocity : } {v}_{y} = {v}_{i} \cdot \sin \alpha$

${v}_{y} = \frac{5}{8} \cdot \sin \left(\frac{5 \pi}{12}\right)$

${v}_{y} = 0 , 60 \text{ m/s}$

$\text{x-component of velocity at any time:} {v}_{x} = {v}_{i} \cdot \cos \alpha$
$\text{y-component of velocity at any time:} {v}_{y} = {v}_{i} \cdot t \cdot \sin \alpha - g \cdot t$

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

${x}_{m} = \frac{{\left(\frac{5}{8}\right)}^{2} \cdot \sin 2 \cdot \frac{5 \pi}{12}}{9 , 81}$

${x}_{m} = \frac{25 \cdot \sin \left(\frac{5 \pi}{6}\right)}{64 \cdot 9 , 81} = \frac{25 \cdot 0 , 5}{64 \cdot 9 , 81}$

${x}_{m} = 0 , 02 \text{ } m$

$\text{for time to maximum height : } t = \frac{{v}_{i} \cdot \sin \alpha}{g}$

$t = \frac{5 \cdot \sin \left(\frac{5 \pi}{12}\right)}{8 \cdot 9 , 81} = \frac{4 , 8296291314}{78 , 48}$

$t = 0 , 06 \text{ s}$