Aball is thrown from a height of 110 feet with an initial downward velocity of 6/fts. the ball's height h (in feet) after t seconds is given by the following: h=110-6t-16t^2 how long after the ball is thrown does it hit the ground? round your answer(s) to the nearest hundredth.

Rearrange equation.h=-16t^2-6t+110 h = 0 at the ground. divide both sides of the equation by (-2) to yield: 0 = 8t^2+3t - 55 and use quadratic equation h=-b±√(b2 - 4ac)                                                    2awhere a = 8, b= 3, c=-55 Substitute:  [-3 ± √(9 - 4*8*(-55))] / (16) = [6 ± √(36 + 1760)] / (16) = [6 ± √(1796)] /16 = (6 ± 42.3792)/16.  Time can only be positive for this problem...so discard negative answer. = 48.3792/16 = 3.02 seconds