The height of the ball at any time t is given by
[tex]h(t)=-16t^2+40t+1.5[/tex]This is a quadratic equation, which attains its maximum value at time:
[tex]t=\frac{-b}{2a}[/tex]In the given equation, a = -16 and b = 40. substitute these values in the formula:
[tex]t=\frac{-40}{-16\times2}=\frac{-40}{-32}=\frac{5}{4}[/tex]Therefore, the ball attains its maximum height at t=5/4 seconds which is given below:
[tex]\begin{gathered} h(\frac{5}{4})=-16(\frac{5}{4})^2+40(\frac{5}{4})+1.5 \\ =-25+50+1.5 \\ =26.5 \end{gathered}[/tex]Thus, the maximum height attained by the ball is 26.5 feet.
