Answer :
a) For this question we set h=59 and solve for t, in order to do so we use the general formula for second-degree equations:
[tex]\begin{gathered} t=\frac{-124\pm\sqrt[]{124^2-4(-16)(-40)}}{2(-16)} \\ t=\frac{-124\pm113.21}{-32} \end{gathered}[/tex]The height of the object will be 59 feet at t=7.41 seconds and t=0.34 seconds.
b) When the object reaches the ground, h=0 therefore:
[tex]0=-16t^{2}+124t+19[/tex]Solving for t we get:
[tex]\begin{gathered} t=\frac{-124\pm\sqrt[]{124^2-4(-16)(19)}}{2(-16)} \\ t=\frac{-124\pm\sqrt[]{16592}}{-32}=\frac{-124\pm128.81}{-32} \end{gathered}[/tex]Therefore, since t cannot be negative the solution is t=7.9 seconds.