Question

A ball is thrown vertically upward. After t seconds, its height h (in feet) is given by the functionh (t) = 441 – 161². What is the maximum height that the ball will reach?Do not round your answer.Height:

Answer 1

Answer:

Height: 30.25 feet.

Explanation:

The equation of the ball's path is given as:

$h(t)=44t-16t^2$

To find the maximum point, first, find the derivative of the function:

$h^{\prime}(t)=44-32t$

Next, set the derivative equal to 0 and solve for t:

$\begin{gathered} 44-32t=0 \\ \implies44=32t \\ t=\frac{44}{32} \\ t=1.375 \end{gathered}$

The maximum height occurs when the time, t=1.375 seconds.

Next, substitute it into h(t):

$\begin{gathered} h(t)=44t-16t^2 \\ =44(1.375)-16(1.375)^2 \\ =60.5-30.25 \\ =30.25\text{ feet} \end{gathered}$

The maximum height, h(t) that the ball will reach is 30.25 feet.