A ball is thrown from an initial height of 3 meters with an initial upward velocity of 20m/s. The ball's height (in meters) after seconds is given by the following h=3+20t-5t^2. Find all values of t for which the ball's height is 13 meters.
+128475 -5t^2 + 20t + 3 = 13 subtract 13 from both sides
-5t^2 + 20t - 10 = 0 divide through by -5
t^2 - 4t + 2 = 0
Using the quadratic formula
t = 4 ± sqrt [ 4^2 - 4 (1) (2) ] 4 ± sqrt [ 8] 4 ± 2sqrt (2)
_____________________ = ___________ = __________ =
2 * 1 2 2
2 ± sqrt (2) = about .59 sec and about 3.4 sec
