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y = –16 t 2 + 24.6 t + 6.5

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y = –16 t 2 + 24.6 t + 6.5

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  1. An athlete on a track team throws a shot put. The height y of the shot put in feet t seconds after it is thrown is modeled by y = –16t2 + 24.6t+ 6.5. The horizontal distance x in between the athlete and the shot put is modeled by x = 29.3t. To the nearest foot, how far does the shot put land from the athlete?

  2. Step 1 Use the first equation to determinehow long it will take the shot put to hit the ground. Set the height of the shot put equal to 0 feet, and the use the quadratic formula to solve for t. y = –16t2 + 24.6t + 6.5 0 = –16t2 + 24.6t + 6.5 Set y equal to 0. Use the Quadratic Formula. Substitute –16 for a, 24.6 for b, and 6.5 for c.

  3. Simplify. The time cannot be negative, so the shot put hits the ground about 1.8 seconds after it is released.

  4. Step 2 Find the horizontal distance that the shot put will have traveled in this time. x = 29.3t x ≈ 29.3(1.77) Substitute 1.77 for t. x ≈ 51.86 Simplify. x ≈ 52 The shot put will have traveled a horizontal distance of about 52 feet.

  5. Properties of Solving Quadratic Equations

  6. Properties of Solving Quadratic Equations

  7. Helpful Hint No matter which method you use to solve a quadratic equation, you should get the same answer.

  8. Lesson Quiz: Part I Find the zeros of each function by using the Quadratic Formula. 1. f(x) = 3x2 – 6x – 5 2. g(x) = 2x2 – 6x + 5 Find the type and member of solutions for each equation. 3. x2 – 14x + 50 4. x2 – 14x + 48 2 distinct real 2 distinct nonreal complex

  9. Lesson Quiz: Part II 5. A pebble is tossed from the top of a cliff. The pebble’s height is given by y(t) = –16t2+ 200, where t is the time in seconds. Its horizontal distance in feet from the base of the cliff is given by d(t) = 5t. How far will the pebble be from the base of the cliff when it hits the ground? about 18 ft

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