A ball 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