Question

A ball is thrown vertically upward with an initial speed of 80 f/s. Its height after t seconds is given by h=80t−16t^2 a)how high does the ball go and when does the ball hit the ground?

Answer 1

max height of the ball is the second coordinate of the vertex of the parabola. the first coordinate is \(-\frac{b}{2a}\) in your case it is \(-\frac{80}{2\times -16}=\frac{5}{2}\)
to find the actual height, replace \(t\) by \(\frac{5}{2}\) and see what you get