A ball is thrown from a height of 80 feet with an initial downward velocity of 4/fts. The ball's height h (in feet) after t seconds is given by the following.

h-80-4t-16t^2

How long after the ball is thrown does it hit the ground? Round your answer (s) to the nearest hundredth.

Alright, so we have h=80-4t-16t^2. We have to find when h=0. We can use the quadratic formula!

(use x as t)

x = (4+-sqrt (16 - (-64*80)) / 32 = (4+-sqrt (5136)) / 32, and since 4-sqrt (5136) is clearly negative that doesn't work, so we have (4+sqrt (5136)) / 32, which is approximately 2.36