# A projectile is shot from the ground at an angle of pi/8  and a speed of 1 /12 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?

Apr 27, 2018

$V e l o c i t {y}_{Y} = \frac{1}{12} \cdot \sin \left(\frac{\pi}{8}\right) = 0.0319 m {s}^{-} 1$
$V e l o c i t {y}_{X} = \frac{1}{12} \cdot \cos \left(\frac{\pi}{8}\right) = 0.0770 m {s}^{-} 1$
Considering vertical movement, since at maximum height final velocity is $0 m {s}^{-} 1$.

${U}_{Y} = 0.0319 m {s}^{-} 1$ ${a}_{Y} = - 9.81 m {s}^{-} 2$ ${V}_{Y} = 0 m {s}^{-} 1$
We use $V = U + a T$,
$0 = 0.0319 + \left(- 9.81\right) T$
$T = 0.00325 s$
Since horizontal component remains unchanged,
Horizontal Distance = $0.0770 \cdot 0.00325 = 2.50$ x ${10}^{-} 4 m$