Question 215013
Note: I'm going to use the variable 'x' instead of 't'. 



{{{y= 144 + 128x- 16x^2}}} Start with the given equation.



{{{y=-16x^2+128x+144}}} Rearrange the terms.



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=-16x^2+128x+144}}}, we can see that {{{a=-16}}}, {{{b=128}}}, and {{{c=144}}}.



{{{x=(-(128))/(2(-16))}}} Plug in {{{a=-16}}} and {{{b=128}}}.



{{{x=(-128)/(-32)}}} Multiply 2 and {{{-16}}} to get {{{-32}}}.



{{{x=4}}} Divide.



So the x-coordinate of the vertex is {{{x=4}}}. Note: this means that the axis of symmetry is also {{{x=4}}}.



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



{{{y=-16x^2+128x+144}}} Start with the given equation.



{{{y=-16(4)^2+128(4)+144}}} Plug in {{{x=4}}}.



{{{y=-16(16)+128(4)+144}}} Square {{{4}}} to get {{{16}}}.



{{{y=-256+128(4)+144}}} Multiply {{{-16}}} and {{{16}}} to get {{{-256}}}.



{{{y=-256+512+144}}} Multiply {{{128}}} and {{{4}}} to get {{{512}}}.



{{{y=400}}} Combine like terms.



So the y-coordinate of the vertex is {{{y=400}}}.



So the vertex is *[Tex \LARGE \left(4,400\right)].



This means that the max height is 400 feet (since the max/min is the y coordinate of the vertex). This max height occurs when {{{t=4}}} (since x=4 at the vertex and I used 'x' instead of 't')