the path of a rocket fired during a fireworks display is given by the equation s(t)=64t-16t^2 (t to the second power) where t is the time, in seconds, and s is the height, in feet. what is the maximum height, in feet, the rocket will reach? In how many seconds will the rocket hit the ground?

Answer 1

This is a parabola. Zeroes (that is s(t) is zero ) when t=0, and t=4.

So the max height due to symettry of the parabola occurs at t=2.