Solution:
Given:
[tex]\begin{gathered} t=\frac{\sqrt[]{d}}{2} \\ t=1.12s \end{gathered}[/tex]Hence, the vertical distance is;
[tex]\begin{gathered} t=\frac{\sqrt[]{d}}{2} \\ 1.12=\frac{\sqrt[]{d}}{2} \\ \text{Cross multiplying the equation,} \\ \sqrt[]{d}=1.12\times2 \\ \sqrt[]{d}=2.24 \\ \\ \text{Taking the square of both sides,} \\ d=2.24^2 \\ d=5.0176 \\ \\ To\text{ the nearest tenth,} \\ d=5.0\text{feet} \end{gathered}[/tex]Therefore, the vertical distance of his jump to the nearest tenth is 5.0 feet