Answer
The maximum height is $25$ feet.
Work Step by Step
We are given the equation representing the height ($y=40t-16t^2$) and have to maximize it.
In other words, find the maximum value of a quadratic polynomial. We can do this by finding the vertex of $y$ using the methods outlined in the chapter; mainly, the vertex of a quadratic polynomial of form $ax^2+bx+c$ being the point $(h, k)$ where $h=-\frac{b}{2a}$ and $k=f(h)$.
In our case, $a=-16, b=40, c=0$ giving us $h=\frac{5}{4}$ and $k=f(\frac{5}{4}) = 25.$ Hence, we know that $y$ attains a maximum of $25$ when $t=\frac{5}{4}$.
Interpreting the results in the context of the problem, this tells us that the ball attains a maximum height of $25$ feet after $\frac{5}{4}$ seconds.
Below is a graph representing the height of the ball. The y-axis represents the height while the x-axis represents time.
