Solution
Step 1
Write the distance equation.
[tex]\text{s = -16t}^2+\text{ 160t}[/tex]Step 2
At maximum height , v = 0
[tex]\begin{gathered} v\text{ = }\frac{ds}{dt} \\ s\text{ }=\text{ -16t}^2\text{ + 160} \\ \frac{ds}{dt}\text{ = 2}\times-16t\text{ + 160} \\ \frac{ds}{dt}\text{ = -32t + 160} \\ -32t\text{ + 160 = 0} \\ 32t\text{ = 160} \\ t\text{ = }\frac{160}{32} \\ \text{t = 5 seconds} \end{gathered}[/tex]At maximum height, t = 5
Step 3
Substitute x = 5 to find the maximum height
[tex]\begin{gathered} \text{s = -16t}^2\text{ + 160t} \\ Maximum\text{ height = -16}\times5^2\text{ + 160}\times\text{ 5} \\ \text{= -16 }\times\text{ 25 + 800} \\ \text{= -400 + 800} \\ =\text{ 400 feet} \end{gathered}[/tex]Final answer
Maximum height = 400 feet
