# How do I represent this in terms of m and x

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?

• i think you just proved that $E$ is independent of $x$... – gt6989b Mar 1 '16 at 3:39
• the "in terms of m and x" then i just meant as a concept check? ...which apparently I've failed :[ – FakeBrain Mar 1 '16 at 3:43
• @ThisIsNotAnId, then $v=h'=1-32t$ should actually be $v=h'=-32t$ ? – FakeBrain Mar 1 '16 at 4:07

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.