I’m not sure if I’m reading this correctly, but I think you meant to write something like…

[math]v=gt[/math]

[math]x(0)=0[/math]

This? If not leave something in the comments to clarify.

The best way to approach this is with good old calculus. You can note that, assuming there is only one dimension, velocity can be expressed as.

[math]v=\frac{dx}{dt}[/math]

Meaning that our original expression can be written as the differential equation

[math]\frac{dx}{dt}=gt[/math]

…”Differential equation? Oh, no!” I hear you say (since I assume you aren’t familiar). It’s really a differential equation by technicality, this is just a simple calculus problem.

Multiply both sides by the differential [math]dt[/math] to get

[math]\frac{dx}{dt}dt=gtdt[/math]

Which can also be written as,

[math]dx=gtdt[/math]

I can almost hear the little guy screaming “integrate me”

[math]\int dx=\int gtdt[/math]

This yields.

[math]x=\frac{gt^2}{2}+C[/math]

So, now that we have that, all we have to do is fix this a bit with our initial condition.

We know that [math]x(0)=0[/math] so…

[math]x(0)=\frac{g(0)^2}{2}+C=0,C=0[/math]

So our equation is now.

[math]x=\frac{gt^2}{2}[/math]

Almost done, now we just plug in our time elapsed, which is 1 second.

[math]x=\frac{g(1)^2}{2}=\frac{g}{2}[/math]

It looks like our units were feet, so [math]x=16.1ft[/math]

There you go!