To find the time that the rock is in the air, we need to find the times when f(t)=0:
[tex]0=-1.9t^2+38t[/tex]We factor this equation as follows:
[tex]0=t(-1.9t+38)[/tex]And now we apply zero product property, if ab=0, then a=0 or b=0.
In this case:
[tex]\begin{gathered} t=0 \\ or \\ -1.9t+38=0 \end{gathered}[/tex]Since t=0 is the stating time, we only take the second equation, and solve for the time:
[tex]\begin{gathered} -1.9t+38=0 \\ -1.9t=-38 \\ t=\frac{-38}{-1.9} \\ t=20 \end{gathered}[/tex]Answer: 20 seconds
