Question

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?

Answer 1

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.