SOLUTION
Now the ground is assumed to be 0, so we have that
[tex]y(t)=0[/tex]So, that means we have
[tex]\begin{gathered} y(t)=-16t^2+100 \\ 0=-16t^2+100 \\ 16t^2=100 \\ t^2=\frac{100}{16} \\ t=\sqrt{\frac{100}{16}} \\ t=\frac{10}{4} \\ t=2.5\text{ seconds } \end{gathered}[/tex]Now, we have found t, which is how long it takes to get to the ground, you can plug it into x(t) to find the horizontal distance travelled, we have
[tex]\begin{gathered} x(t)=8t \\ x(2.5)=8\times2.5 \\ =20\text{ feet } \end{gathered}[/tex]Hence it takes her 2.5 seconds to reach the ground
And she is 20 feet away from the cliff