The equation y = -4.9t^2 + 42t + 18.9 describes the height (in meters) of a ball tossed up in the air at 42 meters per second from a height of 18.9 meters from the ground, as a function of time in seconds. In how many seconds will the ball reach a height of 12 feet?.
+128475 12 ft = 12 (.3048) = 3.6576 m
3.6576 = -4.9t^2 + 42t + 18.9
-4.9t^2 + 42t + 15.2424 = 0
Using the quadratic formula, the ball reaches 12 ft at
-42 - sqrt ( 42^2 - 4 (-4.9)(15.2424) ) -42 - sqrt (2062.75104)
__________________________ = ___________________ ≈ 8.92 sec
2 * -4.9 -9.8
