Quadratic Applications

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Quadratic Applications · Grade 9 · Worksheet 1

1. A parabolic arch bridge spans a river with its vertex at the water's surface. The arch is modeled by the quadratic function f(x) = -0.02x² + 12, where x represents the horizontal distance from the center of the arch in meters, and f(x) represents the height above the water in meters. A boat needs to pass under the bridge with a clearance of at least 2 meters. What is the maximum width of the boat that can pass through the center of the arch while maintaining this clearance? Answer: 20 Solution: f(x) = -0.02x² + 12 - x = horizontal distance from the center (in meters) - f(x) = height above the water (in meters) - Vertex is at (0, 12), meaning at the center (x=0), the arch is 12 m above water.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Understand the problem** We have the arch equation: f(x) = -0.02x² + 12 Here: - x = horizontal distance from the center (in meters) - f(x) = height above the water (in meters) - Vertex is at (0, 12), meaning at the center (x=0), the arch is 12 m above water. The boat needs 2 m clearance above water. So the available height for the boat is: 12 m - 2 m = 10 m above water. We need to find the horizontal distance x from the center where the arch height is exactly 10 m. The width of the boat will be 2x (since it's symmetric). --- **Step 2: Set up the equation for height = 10 m** We set f(x) = 10: -0.02x² + 12 = 10 --- **Step 3: Solve for x²** Subtract 12 from both sides: -0.02x² = 10 - 12 -0.02x² = -2 Divide both sides by -0.02: x² = (-2) / (-0.02) x² = 2 / 0.02 x² = 100 --- **Step 4: Solve for x** x = sqrt(100) = 10 (x = -10 is the other side, but we take the positive one for half-width.) --- **Step 5: Find the total width available for the boat** The boat passes through the center, so the total width between the two points where the arch height is 10 m is: 2 * x = 2 * 10 = 20 meters. --- **Step 6: Interpret the result** The maximum width of the boat that can pass through the center while maintaining at least 2 m clearance is **20 meters**. --- **Final answer:** 20

2. Nova throws a ball straight upward from a platform 6 meters above the ground. The height of the ball above the ground, h(t) in meters, after t seconds is given by the quadratic function h(t) = -5t² + 16t + 6. The graph of this function is a parabola opening downward, with the vertex representing the maximum height of the ball. After how many seconds does the ball reach its maximum height, and what is that maximum height? Answer: 1.6 seconds; 18.8 meters Solution: The height function is h(t) = -5t² + 16t + 6. This is a quadratic in standard form where a = -5, b = 16, c = 6. The ball reaches its maximum height at the vertex of the parabola.
   Full step-by-step solution

   Step 1: The height function is h(t) = -5t² + 16t + 6. This is a quadratic in standard form where a = -5, b = 16, c = 6. Step 2: The ball reaches its maximum height at the vertex of the parabola. The time at the vertex is given by the axis of symmetry formula: t = -b / (2a). Step 3: Substitute the values: t = -16 / (2 * -5) = -16 / -10 = 1.6 seconds. Step 4: To find the maximum height, substitute t = 1.6 into the height function: h(1.6) = -5(1.6)² + 16(1.6) + 6 Step 5: Calculate (1.6)² = 2.56. Then -5 * 2.56 = -12.8. Step 6: 16 * 1.6 = 25.6. Step 7: Now add: h(1.6) = -12.8 + 25.6 + 6 = 12.8 + 6 = 18.8 meters. Step 8: Therefore, the ball reaches its maximum height of 18.8 meters after 1.6 seconds. Answer: 1.6 seconds; 18.8 meters.

3. Emma throws a ball straight upward from a height of 1 meter above the ground with an initial velocity of 15 m/s. The height of the ball above the ground after t seconds is modeled by the quadratic function h(t) = -5t² + 15t + 1, where h(t) is in meters. How many seconds after the throw does the ball hit the ground? Round your answer to the nearest tenth of a second. Answer: 3.1 Solution: When the ball hits the ground, its height is 0 meters. So set h(t) = 0: -5t² + 15t + 1 = 0. Multiply both sides by -1 to make the leading coefficient positive: 5t² - 15t - 1 = 0.
   Full step-by-step solution

   Step 1: When the ball hits the ground, its height is 0 meters. So set h(t) = 0: -5t² + 15t + 1 = 0. Step 2: Multiply both sides by -1 to make the leading coefficient positive: 5t² - 15t - 1 = 0. Step 3: Use the quadratic formula t = (-b ± sqrt(b² - 4ac)) / (2a) with a = 5, b = -15, c = -1. Step 4: Compute the discriminant: b² - 4ac = (-15)² - 4(5)(-1) = 225 + 20 = 245. Step 5: sqrt(245) = sqrt(49 * 5) = 7*sqrt(5) ≈ 7 * 2.236 = 15.652. Step 6: t = (15 ± 15.652) / (2 * 5) = (15 ± 15.652) / 10. Step 7: Two possible times: t = (15 + 15.652) / 10 = 30.652 / 10 = 3.0652 seconds, and t = (15 - 15.652) / 10 = -0.652 / 10 = -0.0652 seconds. Step 8: Since time cannot be negative, the ball hits the ground at t ≈ 3.0652 seconds. Rounded to the nearest tenth, t = 3.1 seconds. The answer is 3.1.

4. Isabella kicks a soccer ball that follows the path h(t) = -5t² + 15t + 10, where h is height in meters and t is time in seconds. What is the maximum height of the ball? Answer: 21.25 Solution: The height function is h(t) = -5t² + 15t + 10 Find the time when maximum height occurs using t = -b/(2a) t = -15/(2×-5) = -15/-10 = 1.5 seconds Substitute t = 1.5 into the height function: h(1.5) = -5(1.5)² + 15(1.5) + 10 Calculate (1.5)² = 2.25 h(1.5) = -5×2.25 + 15×1.5 + 10 = -11.25 + 22.5 +…
   Full step-by-step solution

   Step 1: The height function is h(t) = -5t² + 15t + 10 Step 2: Find the time when maximum height occurs using t = -b/(2a) Step 3: t = -15/(2×-5) = -15/-10 = 1.5 seconds Step 4: Substitute t = 1.5 into the height function: h(1.5) = -5(1.5)² + 15(1.5) + 10 Step 5: Calculate (1.5)² = 2.25 Step 6: h(1.5) = -5×2.25 + 15×1.5 + 10 = -11.25 + 22.5 + 10 Step 7: Combine terms: -11.25 + 22.5 = 11.25, then 11.25 + 10 = 21.25 Step 8: The maximum height is 21.25 meters

5. A parabolic arch bridge spans a river with its vertex at the highest point of the arch, 15 meters above the water. The arch is modeled by a quadratic function in vertex form: y = a(x - h)² + k, where (h, k) is the vertex. The arch meets the water at two points that are 50 meters apart horizontally. If the left support is at the origin (0, 0) on a coordinate plane, what is the equation of the parabola? Answer: y = -0.024x² + 1.2x Solution: Identify the vertex. The highest point is at the vertex, which is halfway between the two supports. Since the supports are 50 meters apart and the left support is at (0, 0), the vertex is at (25, 15).
   Full step-by-step solution

   Step 1: Identify the vertex. The highest point is at the vertex, which is halfway between the two supports. Since the supports are 50 meters apart and the left support is at (0, 0), the vertex is at (25, 15). Step 2: Write the vertex form equation: y = a(x - 25)² + 15. Step 3: Use a known point to find 'a'. The left support is at (0, 0), so substitute x = 0 and y = 0 into the equation: 0 = a(0 - 25)² + 15. Step 4: Simplify: 0 = a(625) + 15. Step 5: Solve for 'a': 625a = -15, so a = -15/625 = -0.024. Step 6: Write the final equation: y = -0.024(x - 25)² + 15. Step 7: Expand to standard form if needed: y = -0.024(x² - 50x + 625) + 15 = -0.024x² + 1.2x - 15 + 15 = -0.024x² + 1.2x. The equation of the parabola is y = -0.024x² + 1.2x.