A projectile is shot from the ground at an angle of pi/6  and a speed of 5 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 14, 2016

$x = 1 , 08 \text{ } m$
$y = 0 , 319 \text{ } m$

Explanation:

${t}_{e} = {v}_{i} \cdot \sin \frac{\alpha}{g} = \frac{5 \cdot 0 , 5}{9 , 81} = 0 , 25 s$
$x = {v}_{i} \cdot t \cdot \cos \alpha$

$x = 5 \cdot 0 , 25 \cdot 0 , 866$
$x = 1 , 08 \text{ } m$

$y = {v}_{i} \cdot t \cdot \sin \alpha - \frac{1}{2.} g . {t}^{2}$
$y = 5 \cdot 0 , 25.0 , 5 - \frac{1}{2} \cdot 9 , 81 \cdot {\left(0 , 25\right)}^{2}$
$y = 0 , 625 - 0 , 306 = 0 , 319$
$y = 0 , 319 \text{ } m$