I'll assume what you intended to type was [tex]h(t)=-8t^2+40t[/tex], otherwise the context of the problem makes little sense.
As the question in under Mathematics and not Physics, I'll do my best to stick to mathematical and not physical description of the solution.
We're given a function that describes the position of an object, this function is quadratic hence a parabola. In this case, the vertex of the parabola is the highest point of motion, so we must find the vertex of the parabola [tex]h(t)=-8t^2+40t[/tex].
The vertex of a parabola is a maxima, a parabola has only one maxima. Given an arbitrary function of an arbitrary variable ([tex]s[/tex]) the maxima is found by solving for [tex]s[/tex] in the equation:
[tex] \frac{d}{ds}[f(s)]=0[/tex]
In this problem [tex]f(s)=f(t)=-8t^2+40t[/tex].
Lets find the maxima/vertex by solving for [tex]t[/tex] in the following:
[tex] \frac{d}{dt}(-8t^2+40t)=-16t+40=0[/tex]
[tex]t= \frac{-40}{-16}= \frac{5}{2}=2.5 [/tex]
This means it takes 2.5 seconds for the object to reach its highest point (vertex/maxima of the parabola).
In order to get the ordered pair that represents the highest point we plug in [tex]t=2.5[/tex] in the function:
[tex]h(2.5)=-8(2.5)^2+40(2.5)=50[/tex]
This means that ordered pair that represents the highest point (vertex/maxima) is (2.5,50).
