Find $\frac{\partial p}{\partial t}$ in terms of $p$ If $\left. p = -4.9t^2\ \middle|\ t \geq 0 \right.$, then $\frac{\partial p}{\partial t} = -9.8t$. How do I find $\frac{\partial p}{\partial t}$ in terms of $p$ rather than $t$ (i.e. velocity in terms of position)? Solving for $t$ in the first equation produces a square-root of a negative.

It is clear that $p\le 0$, so there is no square root of a negative here. We have
$$t=\sqrt{\frac{-p}{4.9}}.$$
(Note we took the non-negative square root, since $t \ge 0$.)
Remark: Presumably you have dropped an object from some height, on an airless planet where the acceleration due to gravity is remarkably like Earth's.
You chose $p=0$ as the coordinate of the drop point. That means that the position at time $t\gt 0$ is negative. Perfectly fine. But if it interferes with the intuition, let $p=0$ at ground level. 
