Respuesta :
[tex]\begin{gathered} \text{ One way of solving the maximum height of the ball with the given equation is} \\ \text{getting the vertex of the function of time since it is a quadratic function given by} \\ x=-\frac{b}{2a},\text{ and substitute time t to the original function} \\ \text{ The other method of finding the maximum height is getting} \\ \text{ the first derivative of the function},\text{ and equate it to first derivative to zero} \\ \text{Solving the maximum height using vertex of the function} \\ \text{ Convert }x=-\frac{b}{2a}\text{ to a function of time we get} \\ t=-\frac{80}{2(-16)}=-\frac{80}{-32} \\ t=2.5\text{seconds} \\ \text{Substitute} \\ s(t)=-16t^2+80t+6 \\ s(2.5)=-16(2.5)^2+80(2.5)+6 \\ s(2.5)=106ft \\ \text{Solving using the first derivative of the function} \\ s(t)=-16t^2+80t+6 \\ s^{\prime}(t)=(2)(-16)t+80+0 \\ s^{\prime}(t)=-32t+80 \\ -32t+80=0 \\ -32t=-80 \\ \frac{-32t}{-32}=\frac{-80}{-32} \\ t=2.5\text{seconds} \\ \text{ Substitute as usual and we get the same result.} \end{gathered}[/tex]
