Question 149906
The maximum height occurs at the vertex of the parabola. So let's find the vertex first





In order to find the vertex, we first need to find the x-coordinate of the vertex.



To find the x-coordinate of the vertex, use this formula: {{{x=(-b)/(2a)}}}.



{{{x=(-b)/(2a)}}} Start with the given formula.



From {{{y=-16t^2+32t}}}, we can see that {{{a=-16}}}, {{{b=32}}}, and {{{c=0}}}.



{{{x=(-(32))/(2(-16))}}} Plug in {{{a=-16}}} and {{{b=32}}}.



{{{x=(-32)/(-32)}}} Multiply 2 and {{{-16}}} to get {{{-32}}}.



{{{x=1}}} Divide.



So the x-coordinate of the vertex is {{{x=1}}}. Note: this means that the axis of symmetry is also {{{x=1}}}.



Now that we know the x-coordinate of the vertex, we can use it to find the y-coordinate of the vertex.



{{{y=-16t^2+32t}}} Start with the given equation.



{{{y=-16t^2+32t}}} Plug in {{{x=1}}}.



{{{y=-16t^2+32t}}} Start with the given equation.



{{{y=-16(1)^2+32(1)}}} Plug in {{{t=1}}}.



{{{y=-16(1)+32(1)}}} Square {{{1}}} to get {{{1}}}.



{{{y=-16+32(1)}}} Multiply {{{-16}}} and {{{1}}} to get {{{-16}}}.



{{{y=-16+32}}} Multiply {{{32}}} and {{{1}}} to get {{{32}}}.



{{{y=16}}} Combine like terms.



So the y-coordinate of the vertex is {{{y=16}}}.



So the vertex is *[Tex \LARGE \left(1,16\right)].



So the max height is 16 feet.




Now if we graph the equation {{{y=-16t^2+32t}}} on the interval *[Tex \LARGE 0\le t \le 2], we can see that the highest point is (1,16). So this confirms our answer.



<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/graph.png" alt="Photobucket - Video and Image Hosting"> Graph of {{{y=-16t^2+32t}}}