jmc

First, we find the maximum height of the ball by maximizing the expression $-25t^2+75t+24$. We will do this by completing the square. Factoring a $-25$ from the first two terms, we have $$-25t^2+75t+24=-25(t^2-3t)+24$$To complete the square, we add and subtract ${\left(-\frac{3}{2}\right)}^2=\frac{9}{4}$ inside the parentheses to get $$\begin{array}{rl}-25(t^2-3t)+24& =-25\left(t^2-3t+\frac{9}{4}-\frac{9}{4}\right)+24\\ & =-25\left({\left(t-\frac{3}{2}\right)}^2-\frac{9}{4}\right)+24\\ & =-25{\left(t-\frac{3}{2}\right)}^2+\frac{225}{4}+\frac{96}{4}\\ & =-25{\left(t-\frac{3}{2}\right)}^2+\frac{321}{4}\end{array}$$Since $-25{\left(t-\frac{3}{2}\right)}^2$ is always non-positive, the maximum value of the expression is achieved when $-25{\left(t-\frac{3}{2}\right)}^2=0$. This happens when $t-\frac{3}{2}=0$. Therefore the height of the ball is at its maximum when $t=\boxed{\frac{3}{2}}$.