Her vertical position is described by y(t). She reach the ground when y(t) = 0 so:
[tex]y(t)=0\\\\-16t^2+100=0\quad|-16t^2\\\\100=16t^2\quad|:16\\\\t^2=6.25\quad|\sqrt{(\dots)}\\\\\boxed{t=2.5}[/tex]
She reach the ground after 2.5 seconds and she will be:
[tex]x(2.5)=8\cdot2.5=\boxed{20}[/tex] (meters) from the cliff.
