Question

If a baseball is projected upward from ground level with an initial velocity of 160 feet per second, then it’s height is a function of time, given by s= -16t^2+160t. What is the maximum height reached by the ball?

Answer 1

Step 1

Write the distance equation.

`\text{s = -16t}^2+\text{ 160t}`

Step 2

At maximum height , v = 0

`\begin{gathered} v\text{ = }(ds)/(dt) \\ s\text{ }=\text{ -16t}^2\text{ + 160} \\ (ds)/(dt)\text{ = 2}*-16t\text{ + 160} \\ (ds)/(dt)\text{ = -32t + 160} \\ -32t\text{ + 160 = 0} \\ 32t\text{ = 160} \\ t\text{ = }(160)/(32) \\ \text{t = 5 seconds} \end{gathered}`

At maximum height, t = 5

Step 3

Substitute x = 5 to find the maximum height

`\begin{gathered} \text{s = -16t}^2\text{ + 160t} \\ Maximum\text{ height = -16}*5^2\text{ + 160}*\text{ 5} \\ \text{= -16 }*\text{ 25 + 800} \\ \text{= -400 + 800} \\ =\text{ 400 feet} \end{gathered}`

Final answer

Maximum height = 400 feet