Question

A ball is thrown vertically upward. after t seconds, it’s height (in feet) is given by the function h (t)=108t-16t^2. What is the maximum height the ball will reach?

Answer 1

Function h(t) gives us the height as a function of time.

$h(t)=108t-16t^2$

Notice that h(t) is a parabola that opens downwards on the plane; therefore, it will have only one critical point and that critical point will be a maximum.

To find the critical point of the function, we need to find its derivative as shown below

$\begin{gathered} h^{\prime}(t)=108-2\cdot16t=108-32t \\ \Rightarrow h^{\prime}(t)=108-32t \end{gathered}$

Then, set h'(t)=0 and solve for t

$\begin{gathered} h^{\prime}(t)=0 \\ \Rightarrow108-32t=0 \\ \Rightarrow t=\frac{108}{32}=\frac{27}{8} \\ \Rightarrow t=\frac{27}{8} \end{gathered}$

Finally, evaluate h(27/8) to obtain the answer

$\begin{gathered} h(\frac{27}{8})=108(\frac{27}{8})-16(\frac{27}{8})^2=\frac{729}{2}-\frac{729}{4}=\frac{729}{4} \\ \Rightarrow h(\frac{27}{8})=\frac{729}{4}=182.25 \\ \end{gathered}$

The answer is 182.25 ft.