a ball is thrown vertically upward. after t seconds, it's height h ( in feet) is given by the function h(t) =80t -16t^2. what is the maximum height that the ball with reach. Do not round answer. Height:

Respuesta :

Solution

For this case we have the following function:

[tex]h(t)=80t-16t^2[/tex]

For this case the maximum occurs when t=0 and we can find the maximum with the first derivate like this:

[tex]80-32t=0[/tex]

Solving for t we got:

[tex]t=\frac{80}{32}=\frac{5}{2}[/tex]

Then the maximum height is:

[tex]h(2.5)=80\cdot2.5-16(2.5)^2=100[/tex]