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Quadratic Application Problems. Algebra Unit 11. Quadratic Application Problems. If an object is tossed into the air… t he path of the object is represented by the equation at 2 + bt + c = h where h is the height after t seconds a is the acceleration due to gravity
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Quadratic Application Problems Algebra Unit 11
Quadratic Application Problems If an object is tossed into the air… • the path of the object is represented by the equation at2+ bt+ c = h • where h is the height after tseconds • ais the acceleration due to gravity • bis the initial velocity • cis the initial height
modeling A ball is thrown straight up, from 3 ft above the ground, with an initial velocity of 14 ft/sec. It’s the acceleration due to gravity is -5 ft/sec. Use the equation at2 + bt + c = h Use any method to solve for t (factoring, Quadratic Formula, square roots). A) Write the quadratic equation representing this scenario when h= 0. B) Find the roots (solutions) for this quadratic equation. -15 -1 15 14
modeling A ball is thrown straight up, from 3 ft above the ground, with an initial velocity of 14 ft/sec. It’s the acceleration due to gravity is -5 ft/sec. Substitute 2 seconds for t and solve for h. Recall the roots (solutions) you already found. Which one makes sense? C) How high will the ball be after 2 seconds? h = 11 feet after 2 seconds D) How long will it take for the ball to hit the ground?
y x modeling Find the vertex of the parabola. Find the y – intercept. Then reflect that point across the axis of symmetry. Plot the x – intercepts . Draw a smooth curve connecting the points. E) Graph Thus the vertex is (1.4 ,12.8). Graph applied to problem means that the graph starts at (0,3). y – intercept (–0.2 ,0) (3,0) x – intercepts
Modeling F) Using the graph, what is the maximum height the ball reaches? The maximum height the ball reaches is the y– value of the vertex. Since the vertex is (1.4 ,12.8), the ball reaches a maximum height of 12.8 feet.
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