From Mechanics

The frisby hits the ground when the \displaystyle \mathbf{k} term is zero.

This gives

\displaystyle \begin{align} & 8t-{{t}^{\ 2}}=0 \\ & \\ & {{t}^{\ 2}}-8t=0 \\ \end{align}

This quadratic equation in \displaystyle t is to be solved. We factorise the equation.

\displaystyle t \left( t-8 \right)=0

giving solutions

\displaystyle t=0 and \displaystyle t=8

The first solution is correct and shows that the frisby is thrown from ground level. We choose the second solution which gives the time when the frisby once more is at ground level.

\displaystyle t=8 \text { s}