I have a object in free fall with $g$ = acceleration, $y$ is the position above the ground and $t$ = time.
I worked out that to find the speed at and $t$ is $dy = g . t$
So to get the position $py$ at time t is $py = y + \cfrac{g}{2} t^2$
Then to get the time $t$ it would take to reach a distance $d$ is $t = \sqrt { \cfrac{d}{\cfrac{g}{2}}}$
Now I have a starting speed $s$ so to get position $py$ at $t$ is $py = y + \cfrac{g}{2}t^2 + s.t$
The problem I have is I can not work out how to get $t$ on one side of the equation $py = y + \cfrac{g}{2}t^2 + s.t$ where I have position $py$, starting position $y$, acceleration $g$, and starting speed s
I must be missing something basic or just a brain fry, so hope someone can show me how to get the t out to one side. Thanks.
You have a quadratic in $t$, so you have two options; you can use the quadratic formula or you can complete the square. The quadratic formula is a little more straightforward, though, and it says that for an equation of the form $0=c+bt+at^2,$ $$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ Applying this to your equation gives
$$ \begin{eqnarray*} 0&=&(y-py)+st+\frac{g}{2}t^2\\ t&=&\frac{-s\pm\sqrt{s^2-4(y-py)\left(\frac{g}{2}\right)}}{2(y-py)}\\ \end{eqnarray*}$$
One quick note: The conventions are that gravity acts in the negative direction of the y-axis, so your quadratic term should be $-\frac{g}{2}t^2.$