we have the function
[tex]s(t)=-16t^2+20t+2[/tex]this function represents a vertical parabola open downward
the vertex represents a maximum
Convert the given function to vertex form
so
y=a(x-h)^2+k
where
(h,k) is the vertex
step 1
Factor of -16
[tex]s(t)=-16(t^2-\frac{20t}{16})+2[/tex]step 2
Complete the square
[tex]s(t)=-16(t^2-\frac{20t}{16}+\frac{20^2}{32^2}-\frac{20^2}{32^2})+2[/tex][tex]s(t)=-16(t^2-\frac{20t}{16}+\frac{20^2}{32^2})+2+\frac{16\cdot20^2}{32^2}[/tex]simplify
[tex]s(t)=-16(t^2-\frac{20t}{16}+\frac{20^2}{32^2})+2+\frac{25}{4}[/tex][tex]s(t)=-16(t^2-\frac{20t}{16}+\frac{20^2}{32^2})+\frac{33}{4}[/tex]Rewrite as perfect squares
[tex]s(t)=-16(t^{}-\frac{20}{32})+\frac{33}{4}[/tex]simplify
[tex]s(t)=-16(t^{}-\frac{5}{8})+\frac{33}{4}[/tex]the vertex is the point (5/8,33/4)
therefore
the maximum height is the y-coordinate of the vertex
maximum height is 33/4=8.25 ft
