The components of the ball's position [tex]r[/tex] at time [tex]t[/tex] are
[tex]\begin{cases}r_x(t)=v_{0x}t\\v_y(t)=30.0-4.9t^2\end{cases}[/tex]
The ball stops 18.0 m from where it began, so that
[tex]\begin{cases}18.0=v_{0x}t\\0=30.0-4.9t^2\end{cases}[/tex]
From the second equation, we can show that the ball travels for about [tex]t=2.47[/tex] seconds, which means it was initially thrown with a horizontal velocity of
[tex]18.0\,\mathrm m=v_{0x}(2.47\,\mathrm s)\implies v_{0x}=7.29\,\dfrac{\mathrm m}{\mathrm s}[/tex]
