This is the problem:
Let $t$ be the time it takes an object to fall $x$ feet. The kinetic energy of a ball of mass, $m$ dropped vertically $x$ feet is $E = {1 \over 2} m v^2$, where v = $h'$, and $h = x - 16t^2$. Find a formula for E in terms of $m$ and $x$.
So: $$h' = 1 -32t$$ $$v = h' = 1 - 32t$$ which leaves me with the equation: $$E = {1 \over 2} m * (1-32t)^2$$
Here I have $m$ as one of my inputs, but no where is there an $x$, what am I doing wrong?
As gt6989b hinted, $x$ is constant, so its derivative is 0. The kinetic energy is actually independent of initial vertical position as it is a function of speed alone.
Yes the speed of the object will be higher if dropped from a greater height but this will be reflected in the range for the values of $t$ you can pick. Of course, the falling object will have a maximum kinetic energy right before hitting the ground. So, $h^\prime = -32t$. Be sure to calculate and indicate the maximum value $t$ can take.