A model rocket is launched with an initial upward velocity of 54 m/s. The rocket's height h (in meters) after t seconds is given by the following.

Answer:Step-by-step explanation:Let's first put that polynomial in standard form by descending powers of t:[tex]h(t)=-5t^2+54t[/tex]If we are looking for the times where the rocket is at a height of 26 m, we sub in a 26 for h(t) and factor to solve for t.  First we get all the terms on one side of the equals sign and set it equal to zero.[tex]-5t^2+54t-26=0[/tex]That is a second degree polynomial so you can factor it however you are most comfortable to get the values of t are:t = .51 sec and t = 10.29 secSince this is a polynomial, there will be 2 values of t for this time with y being the height and x being the time.  The parabola is symmetrical about the vertex so there are 2 values of t (or x) that equal height (or y).