A ball is thrown from a height of 255 feet with an initial downward velocity of 21/fts. The ball's height h (in feet) after t seconds is given by the following. How long after the ball is thrown does it hit the ground?

Formula is: h = vi · t + 1/2 g · t²

For the final position (on the ground) : h = - 255 ft

vi = - 21 ft/s, g = - 9.81 m/s² = - 32.174 ft/s²

- 255 = - 21 t - 1/2 · 32.174 t²

16.087 t² + 21 t - 255 = 0

t 1/2 = ( - 21 + / - √ (441 + 16,408.74)) / 32.174 =

= (-21 + / - 129.8) / 32.174 = 108.8 / 32.174 = 3.38 s

Answer: The ball hits the ground after 3.38 s.