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 sec
onds 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=<span>-b±√(b2 - 4ac)</span> 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
