jmc

To find the maximum height of the ball is to maximize the expression $-16t^2+80t+21$. We will do this by completing the square. Factoring a $-16$ from the first two terms, we have $$-16t^2+80t+21=-16(t^2-5t)+21$$To complete the square, we add and subtract $(-5/2{)}^2=6.25$ inside the parenthesis to get $$\begin{array}{rl}-16(t^2-5t)+21& =-16(t^2-5t+6.25-6.25)+21\\ & =-16([t-2.5{]}^2-6.25)+21\\ & =-16(t-2.5{)}^2+121\end{array}$$Since $-16(t-2.5{)}^2$ is always non-positive, the maximum value of the expression is achieved when $-16(t-2.5{)}^2=0$, so the maximum value is $0+121=\boxed{121}$ feet.