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.

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.