The answer is very simple. .
[tex]h(t)=-16t^2+6\text{ }[/tex]In this equation the heights values are substituted to determine the times.
For h = 2 feet
[tex]\begin{gathered} 2=-16t^2\text{ + 6 } \\ -16t^2\text{ = 2 - 6} \\ -16t^2\text{ = -4} \end{gathered}[/tex][tex]\begin{gathered} t^2\text{ = }\frac{-4}{-16\text{ }} \\ t^2\text{ = }\frac{1}{4} \end{gathered}[/tex][tex]t\text{ = }\sqrt[]{\frac{1}{4}}\text{ = }0.5\text{ seconds}[/tex]That would be the first time corresponding to the height of 2 feet.
We still need to calculate the time for the other height.
For h = 5 feet
[tex]-16t^2+6\text{ =5}[/tex][tex]\begin{gathered} -16t^2\text{ = 5 -6 } \\ -16t^2\text{ = -1} \\ t^2\text{ = }\frac{1}{16} \\ t\text{ = }\sqrt[]{\frac{1}{16}}\text{ = }0.25\text{ seconds} \end{gathered}[/tex]The answer is: Riley should try to catch the ball between 0.25 and 0.5 seconds.
