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9-4 Quadratic Equations and Projectiles

This text explains quadratic equations and how they relate to projectile motion, specifically in the context of a model rocket launch and a ball thrown into the air. It provides formulas for calculating the height of objects at different times, determining maximum height, and predicting whether an object will clear an obstacle. The examples are solved step-by-step and include graphing and numerical calculations.

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9-4 Quadratic Equations and Projectiles

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  1. 9-4 Quadratic Equations and Projectiles A model rocket is shot at an angle into the air from the launch pad. 1. The height of the rocket when it has traveled horizontally x feet from the launch pad is given by h = - .163x² + 11.43x. A. Graph this equation x= -b 2a y x 0 15 25 35 45 55 70 y 0 135 184 200 184 136 1  200 150 100 50   x= -(11.43) 2(-.163)   x= -11.43 -.326 x 35   x 10 20 30 40 50 60 70 This is the x coordinate of the vertex

  2. B. A 75 ft. tree, 10 ft. from the launch pad, is in the path of the rocket. Will the rocket clear the top of the tree? h = -.163(10)² + 11.43(10) h = -16.3 + 114.3 h = 98 When the rocket is 10ft from the pad it will be 98 ft. above the ground which will clear the 75 ft. tree. Sec. 9-4

  3. 2. The rockets height h at t seconds after launch is given by h = -22.2t² + 133t . -b 2a -133 2(-22.2)  3 x = = a. Graph this equation. b. How high is the rocket in 2 sec.? x 0 1 2 3 4 5 6 y 0  111  177  200  177 110  -1 y From the graph its  175 ft. Using the equation its:  200 150 100 50     h= -22.2t² + 133t -22.2(2)² + 133•2 -88.8 + 266 177.2 ft.   x 1 2 3 4 5 6 Sec. 9-4

  4. General Formula for the Height of a Projectile over Time Let h be the height (in feet) of a projectile launched with an initial velocity v feet per second and an initial height of s feet. Then, after t seconds, h = -16t² + vt + s . Since 16ft ≈ 4.9 meters, if the units are in meters in the formula above, then h = -4.9t² + vt + s .

  5. 1. A ball is thrown from an initial height of 10 feet with an initial velocity of 64 feet per second. a. Write an equation describing the height h in feet of the ball after t seconds. b. How high will the ball be after 3 seconds? c. What is the maximum height of the ball? • h = -16t² + vt + s • h = -16t² + 64t + 10 • b. h = -16(3)² + 64(3) + 10 • h = -144 + 192 + 10 • h = 58 ft • t = - b = -(64) = 2 • 2a 2(-16) • h = -16(2)² + 64(2) + 10 • h = -64 + 128 + 10 • h = 74 ft

  6. 2. An object is dropped from an initial height of 40meters. a. Write a formula describing the height of the object (in meters) after t seconds. b. After how many seconds does the object hit the ground? c. What is the maximum height of the object? • h = -4.9t² + vt + s • h = -4.9t² + 40 • b. h = -4.9t² + 40 • 0 = -4.9t² + 40 • -40 - 40 • -40/-4.9 = t² • √-40/-4.9 = √t² • √8.2 ≈ √t² • 2.9sec. ≈ t • t = - b = 0 = 0 • 2a 2(-4.9) • h = -4.9(0)² + 40 • h = 40m

  7. 3. Suppose a ball is thrown upward with an initial upward velocity of 30 meters per second from an initial height of 10 meters. a Write a formula for the height in meters of the object after t seconds. b. Estimate when the ball is 40 meters high. t = - b = -30 ≈ 3 2a 2(-4.9) • h = -4.9t² + vt + s • h = -4.9t² + 30t + 10 • b. 40 = -4.9t² + 30t + 10 80 ◙ ◙ window 0 < x < 10 0 < y < 80 1.2 4.9 1.2 sec and 4.9 sec 0 10

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