Question

An object is launched at 19.6 m/s from a height of 58.8 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.9t2 +19.6t + 58.8. What is the object's maximum height?

Answer 1

notice that \(\bf h(t) = -4.9t2 +19.6t + 58.8\) is a parabola

notice the leading coefficient is negative, so in the form of \(\bf ax^2+bx+c\) "a" is negative
that means the parabola opens DOWNWARD, with a "hump"

the HIGHEST or LOWEST point in a parabola, is its vertex

for a parabola of that type, you can find the vertex at \(\bf \left(-\cfrac{b}{2a}\quad ,\quad c-\cfrac{b^2}{4a}\right)\)

"x" is usually the "seconds" or time
"y" is the "feet" or height

you only want to know how HIGH it gets, thus you only need the y-coordinate from there, that is \(\bf c-\cfrac{b^2}{4a}\)