1 second after Deon throws it
1) Since the ball goes in the air as described by h(t) = -5t²+t +4, then we can visualize roughly this way:
2) Then we can assume that the moment the ball hits the ground, is one of the roots of that equation. So let's solve it:
![\begin{gathered} h(t)=-5t^2+t+4 \\ -5t^2+t+4=0 \\ \Delta=b^2-4ac\text{ }\Rightarrow\Delta=(1)^2-4(-5)(4)\Rightarrow\Delta=81 \\ x=\frac{-b\pm\sqrt[]{\Delta}}{2a}\Rightarrow x=(-1\pm9)/(2(-5))=1\text{ and -4/5} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/1dre0mliq77rbpt5nzu2cxnpj4c8btxjji.png)
3) So since the negative root of it does not matter to us, since there's no negative time. The ball will hit the ground 1 second after Deon throws it.