Using a Quadratic Equation Problems
A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per
second. The distance
(in feet) of the ball from the ground after
= 112 + 96t - 16t
Complete the table and discuss the interpretation of each point.
After 0 (t) seconds, the ball’s distance from the ground (s) is 112 feet.
because the ball has not been set into motion yet and the building is 112 feet tall.
After 0.5 (t) seconds, the ball’s distance from the ground (s) is 156 feet.
because the ball is initially thrown vertically upward, which would put it above the
building’s height of 112 feet.
After 1 second (t), the ball’s distance from the ground (s) is 192 feet.
because after being thrown vertically, the ball will continue to rise until gravity
takes over and begins to pull it downward.
After 2 seconds (t), the ball’s distance from the ground (s) is 240 feet.