a. At its maximum height, the ball has zero vertical velocity, so
[tex]-\left(28\dfrac{\rm m}{\rm s}\right)^2=2(-g)(y_{\rm max}-17\,\mathrm m)[/tex]
[tex]\implies y_{\rm max}=57\,\mathrm m[/tex]
b. The ball's height in the air [tex]y[/tex] at time [tex]t[/tex] is given according to
[tex]y=17\,\mathrm m+\left(28\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2[/tex]
Solve for [tex]t[/tex] when [tex]y=0[/tex]:
[tex]17+28t-4.9t^2=0\implies t\approx6.3\,\mathrm s[/tex]