Check the picture below.
the ball is higher than the building when y = 0, or namely when s(t) = 0, any other time is above it, and hmmm when it's that?
[tex]\stackrel{s(t)}{0}~~ = ~~-16t^2+64t+400\implies 0=-16(t^2-4t-25) \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{quadratic formula} \\\\ 0=\stackrel{\stackrel{a}{\downarrow }}{1}t^2\stackrel{\stackrel{b}{\downarrow }}{-4}t\stackrel{\stackrel{c}{\downarrow }}{-25} \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a} \\\\\\ t= \cfrac{ - (-4) \pm \sqrt { (-4)^2 -4(1)(-25)}}{2(1)}\implies t=\cfrac{4\pm\sqrt{116}}{2}[/tex]
[tex]t=\cfrac{4\pm 2\sqrt{29}}{2}\implies t= \begin{cases} 2+\sqrt{29}~~\textit{\Large \checkmark}\\ 2-\sqrt{29} \end{cases}~\hfill \boxed{(0~~,~~2\sqrt{29})}[/tex]
notice, we can't use the 2-√29 because is a negative value and "t" is never negati ve, also notice the interval notation, we have parentheses on both sides.
