Given:
[tex]h(t)=-16t^2+104t+56[/tex]
Models the balls height height about the ground.
Required:
To find the maximum height of the ball.
Explanation:
Consider
[tex]\begin{gathered} h(t)=-16t^2+104t+56 \\ a=-16 \\ b=104 \\ c=56 \end{gathered}[/tex][tex]\begin{gathered} t=-\frac{b}{2a} \\ \\ t=-\frac{104}{2(-16)} \\ \\ t=\frac{204}{32} \\ \\ t=3.25 \end{gathered}[/tex]
At t = 3.25 the maximum height is
[tex]\begin{gathered} h(3.25)=-16(3.25)^2+104(t)+56 \\ =225 \end{gathered}[/tex]
Final Answer:
The maximum height of the ball is 225 feet.
