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.
