# A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 1/3 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?

Jan 6, 2016

I found: $0.003 m$

#### Explanation:

Consider the (crappy) diagram:

You get:
$t = \frac{\frac{1}{3} \cdot \sin \left({75}^{\circ}\right)}{9.8} = 0.033 s$ the time to get to the height $h$.
Along $x$ we use this time to find:
${v}_{i x} = \frac{x}{t}$
and:
$x = \frac{1}{3} \cdot \cos \left({75}^{\circ}\right) \cdot 0.033 = 0.003 m$