Question

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?

Answer 1

I found: `0.003m`

Explanation:

Consider the (crappy) diagram:

You get:
`t=(1/3*sin(75^@))/9.8=0.033s` the time to get to the height `h`.
Along `x` we use this time to find:
`v_(ix)=x/t`
and:
`x=1/3*cos(75^@)*0.033=0.003m`