Question 370121
{{{a=dv/dt=-g}}}
{{{v=dx/dt=-gt+v0}}}
{{{x=-(1/2)gt^2+v0*t+x0}}}
Since acceleration is the derivative of velocity, when you integrate you get a term gravity linear in {{{t}}}.
When you integrate velocity to get position, that linear term becomes a quadratuc term in {{{t}}} the 1/2 is brought out during integration.
Hopefully that made sense to you.