from a height of 18.9 meters from the ground, as a function of time in seconds. In how many seconds will the ball hit the ground?
Quadratic formula for ax^2 + bx + c: x = (-b pm sqrt(b^2 - 4ac))/2a
Plug into the formula to get t = 23 seconds
0 ht is when the ball hits the ground
0 = -4.9 t^2 + 42t + 18.9 a = -4.9 b = 42 c = 18.9 for Quadratic Formula:
\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Sub in the values to find t = 9 seconds or - .428 (throw out )
+466 
