Questions: The height above the ground of a stone thrown upwards is given by s(t), where t is measured in seconds. After 2 seconds, the height of the stone is 43 feet above the ground, and after 2.25 seconds, the height of the stone is 54 feet above the ground. Evaluate s(2) and s(2.25), and then find the average velocity of the stone over the time interval [2,2.25]. Evaluate s(2) and s(2.25) s(2)= s(2.25)= (Simplify your answers.) The average velocity of the stone over the time interval [2,2.25] is (Simplify your answer.)
![The height above the ground of a stone thrown upwards is given by s(t), where t is measured in seconds. After 2 seconds, the height of the stone is 43 feet above the ground, and after 2.25 seconds, the height of the stone is 54 feet above the ground. Evaluate s(2) and s(2.25), and then find the average velocity of the stone over the time interval [2,2.25].
Evaluate s(2) and s(2.25)
s(2)=
s(2.25)=
(Simplify your answers.) The average velocity of the stone over the time interval [2,2.25] is
(Simplify your answer.)](https://thumbnail.justsolvely.com/seoQuestionThumb/a19818bf-51e5-4b19-9c04-d054eabb3078.webp)
Transcript text: The height above the ground of a stone thrown upwards is given by $s(t)$, where $t$ is measured in seconds. After 2 seconds, the height of the stone is 43 feet above the ground, and after 2.25 seconds, the height of the stone is 54 feet above the ground. Evaluate $s(2)$ and $s(2.25)$, and then find the average velocity of the stone over the time interval $[2,2.25]$.
Evaluate $s(2)$ and $s(2.25)$
\[
\begin{array}{l}
s(2)=\square \\
s(2.25)=\square
\end{array}
\]
(Simplify your answers.)
The average velocity of the stone over the time interval $[2,2.25]$ is $\square$ $\square$
(Simplify your answer.)
Solution
Was this solution helpful?
Answered
