+1207 The equation \(y = -4.9t^2 + 23.8t\) describes the height (in meters) of a projectile launched from the ground at 23.8 meters per second. In how many seconds will the projectile first reach 28 meters in height?
+37146 The height is given by 'y' set y = to 28
28 = -4.9t^2 + 23.8t now solve for 't'
-4.9t^2 + 23.8t - 28 = 0 Use the quadratic formula (a = -4.9 b = 23.8 c = -28 ) to solve for 't'
\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Note: You will find TWO values of 't'..... the FIRST one is the projectile going UP....the second is the projectile coming DOWN...
