he equation y = -16t^2 + 80t describes the height (in feet) of a projectile launched from the ground at 80 feet per second. At what $t$ will the projectile reach 36 feet in height for the first time? Express your answer as a decimal rounded to the nearest tenth.
Substitute y = 36
36 = -16t^2+80 t now solve for t
16t^2-80t + 36 = 0 use quadratic formula...you will find 2 values for t....one will be the popint it is rising (the first time it reaches 36 ft) and the other will be when it is coming back down
Quadratic formula a = 16 b = - 80 c = 36
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)