Maximum Height Calculation
A ball travels on a parabolic path in which the height (in feet) is given by the expression $ -25t^2+75t+24 $, where $ t $ is the time after launch. At what time is the height of the ball at its maximum?
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Solution
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{align*}
-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{align*}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}} $.
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