From Mechanics

We insert \displaystyle t=0\ \text{s} in the expression for the position vector to find the point where the frisby was thrown from.

\displaystyle \begin{align} & \mathbf{r}=\left( 5\times 0-{{0}^{2}} \right)\mathbf{i}+\left( 7\times 0 \right)\mathbf{j}+0\mathbf{k} \\ & \\ & =0\mathbf{i}+0\mathbf{j}+0\mathbf{k}\ \text{m}\\ \end{align}

Thus the point where the frisby was thrown from is the origin.

Using the result from part b)

\displaystyle \begin{align} & \mathbf{r}=-24\mathbf{i}+56\mathbf{j}+0\mathbf{k}\ \text{m} \end{align}

we get the distance from the origin to where it hits the ground is

\displaystyle \sqrt{{{\left( -24 \right)}^{2}}+{{56}^{2}}}=\sqrt{576+3136}=\sqrt{3712}=60\textrm{.}9\ \text{m}