Question 353196
By thrown into the air, I'm assuming you gave it a velocity in the opposite direction of gravity. 
I'll call that velocity, {{{v[0]}}}.
At time, {{{t=0}}}, the rock is at 30 feet elevation.
The ground is at 0 feet.
.
.
.
{{{y(t)=30+v[0]*t-(1/2)gt^2}}}