Quadratic Equation.docx - Using a Quadratic Equation...

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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) Step-by-Step Algebra Written Explanation of Steps Interpretation 0 112 s (t) = 112 + 96t - 16t 2 s (0) = 112 + 96(0) - 16(0) 2 s (0) = 112 + 0 + 0 s (0) =112 Replace t with 0 Multiply coefficients by value of t Solve At 0 (t) seconds, the ball has not been thrown yet, so it is 112 ft (s) above the ground because the height of the building is 112 feet tall. 0.5 156 s (t) = 112 + 96t - 16t 2 s (.5) = 112 + 96(.5) - 16(.5) 2 s (.5) = 112 + 48 - 4 s (.5) = 156 Replace t with .5 Multiply coefficients by value of t Solve At 0.5 (t) seconds, the ball has been thrown from a height of 112 ft, so it is traveling vertically and is 156 ft (s) above the ground. 1 192 s (t) = 112 + 96t - 16t 2 s (1) = 112 + 96(1) - 16(1) 2 s (1) = 112 + 96 -16 s (1) = 192 Replace t with 1 Multiply coefficients by value of t Solve At 1(t) second, the ball is still traveling vertically at a height of 192 ft (s) but at a slower rate because gravity is gradually pulling it downward. 2 240 s (t) = 112 + 96t - 16t 2 s (2) = 112 + 96(2) - 16(2) 2 s (2) = 112 + 192 - 64 s (2) = 240

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