Gravity is a uniform accleration. So I found the answer by using two of the formulas for uniform acceleration:
[tex]v^{2}=v_{0}^2 + 2ax[/tex]
[tex]v=at+v_0[/tex]
The first lets us find the initial velocity and the second lets us use that to get the time. The first is:
[tex](23\frac{m}{s} )^2=v_0^2+2(-9.8\frac{m}{s^2})(19m) \\ \\ v_0^2=529+372.4\\ \\ v_0=30.02\frac{m}{s}[/tex]
Now using equation 2:
[tex]23\frac{m}{s}=-9.8\frac{m}{s^2}t+30.02\frac{m}{s}[/tex]
and solving this for t:
[tex]23\frac{m}{s} -30.02\frac{m}{s} =-9.8\frac{m}{s^2}t\\\\\frac{-7.02}{-9.8}=0.716s[/tex]
