MATH 111_Quadratic Equation - Discussion.docx - Using a...

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

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