Quadratic Applications

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Answer Key & Explanations

Quadratic Applications · Grade 9 · Worksheet 1

1. Sophia throws a ball straight upward from the top of a 16-meter tall building with an initial velocity of 21 m/s. The height of the ball above the ground is modeled by the quadratic function h(t) = -5t² + 21t + 16, where h is the height in meters and t is the time in seconds after the throw. What is the maximum height the ball reaches above the ground? Answer: 38.05 Solution: The height function is h(t) = -5t² + 21t + 16. Since a = -5 < 0, the parabola opens downward, so the vertex gives the maximum height. The t-coordinate of the vertex is found using t = -b / (2a).
   Full step-by-step solution

   Step 1: The height function is h(t) = -5t² + 21t + 16. Step 2: Since a = -5 < 0, the parabola opens downward, so the vertex gives the maximum height. Step 3: The t-coordinate of the vertex is found using t = -b / (2a). Step 4: Here, b = 21 and a = -5, so t = -21 / (2 × (-5)) = -21 / (-10) = 2.1 seconds. Step 5: Substitute t = 2.1 into the height function to find the maximum height: h(2.1) = -5(2.1)² + 21(2.1) + 16. Step 6: Calculate (2.1)² = 4.41, so -5(4.41) = -22.05. Step 7: Then 21(2.1) = 44.1. Step 8: Add the terms: -22.05 + 44.1 + 16 = 22.05 + 16 = 38.05. Step 9: The maximum height the ball reaches is 38.05 meters.

2. Mere kicks a soccer ball with height h(t) = -16t² + 48t + 4. Find the maximum height of the ball. Answer: 40 Solution: The height function is h(t) = -16t² + 48t + 4. This is a downward-opening parabola, so the maximum height occurs at the vertex.
   Full step-by-step solution

   Step 1: The height function is h(t) = -16t² + 48t + 4. This is a downward-opening parabola, so the maximum height occurs at the vertex. Step 2: The t-coordinate of the vertex is given by t = -b/(2a) = -48/(2×(-16)) = -48/(-32) = 1.5 seconds. Step 3: Substitute t = 1.5 into the height function: h(1.5) = -16(1.5)² + 48(1.5) + 4 Step 4: Calculate (1.5)² = 2.25, so -16 × 2.25 = -36 Step 5: Calculate 48 × 1.5 = 72 Step 6: Combine all terms: -36 + 72 + 4 = 40 Step 7: The maximum height is 40 feet.

3. A ball is thrown upward from a height of 5 meters with an initial velocity of 20 m/s. The height is given by h(t) = -5t² + 20t + 5. Find the maximum height reached by the ball. Answer: 25 Solution: The height function is h(t) = -5t² + 20t + 5. This is a downward-opening parabola (since a = -5 < 0), so the vertex gives the maximum height.
   Full step-by-step solution

   Step 1: The height function is h(t) = -5t² + 20t + 5. This is a downward-opening parabola (since a = -5 < 0), so the vertex gives the maximum height. Step 2: The t-coordinate of the vertex is given by t = -b/(2a) = -20/(2×(-5)) = -20/(-10) = 2 seconds. Step 3: Substitute t = 2 into the height function: h(2) = -5(2)² + 20(2) + 5 = -5(4) + 40 + 5 = -20 + 40 + 5 = 25. Step 4: The maximum height is 25 meters.

4. Matiu kicks a soccer ball whose height is modeled by h(t) = -5t² + 15t + 12. What is the maximum height the ball reaches? Answer: 23.25 Solution: The height function is h(t) = -5t² + 15t + 12. Since the coefficient of t² is negative, the parabola opens downward, and the vertex gives the maximum height. The t-coordinate of the vertex is found using t = -b/(2a).
   Full step-by-step solution

   Step 1: The height function is h(t) = -5t² + 15t + 12. Since the coefficient of t² is negative, the parabola opens downward, and the vertex gives the maximum height. Step 2: The t-coordinate of the vertex is found using t = -b/(2a). Here, a = -5 and b = 15, so t = -15/(2×-5) = -15/-10 = 1.5 seconds. Step 3: Substitute t = 1.5 into the height function: h(1.5) = -5(1.5)² + 15(1.5) + 12 Step 4: Calculate (1.5)² = 2.25, so -5 × 2.25 = -11.25 Step 5: Calculate 15 × 1.5 = 22.5 Step 6: Combine all terms: -11.25 + 22.5 + 12 = 11.25 + 12 = 23.25 Step 7: The maximum height is 23.25 meters.

5. 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.

6. A rocket is launched from ground level with an initial velocity of 128 feet per second. The height h of the rocket after t seconds is given by the function h(t) = -16t² + 128t. After how many seconds will the rocket reach a height of 192 feet? Answer: 2 Solution: Set up the equation using the given height: -16t² + 128t = 192 Move all terms to one side: -16t² + 128t - 192 = 0 Divide the entire equation by -16 to simplify: t² - 8t + 12 = 0 Factor the quadratic: (t - 2)(t - 6) = 0 Solve for t: t = 2 or t = 6 Both values are valid mathematically, but t = 2…
   Full step-by-step solution

   Step 1: Set up the equation using the given height: -16t² + 128t = 192 Step 2: Move all terms to one side: -16t² + 128t - 192 = 0 Step 3: Divide the entire equation by -16 to simplify: t² - 8t + 12 = 0 Step 4: Factor the quadratic: (t - 2)(t - 6) = 0 Step 5: Solve for t: t = 2 or t = 6 Step 6: Both values are valid mathematically, but t = 2 represents when the rocket first reaches 192 feet on its way up, and t = 6 represents when it passes 192 feet again on its way down. The question asks for when it reaches this height, so the first time is t = 2 seconds. The answer is 2.

7. Aroha kicks a soccer ball whose height is given by h(t) = -5t² + 15t + 2. Find the maximum height of the ball. Answer: 13.25 Solution: The height function is h(t) = -5t² + 15t + 2. Since the coefficient of t² is negative, the parabola opens downward, so the vertex gives the maximum height. Find the time t when maximum height occurs using t = -b/(2a).
   Full step-by-step solution

   Step 1: The height function is h(t) = -5t² + 15t + 2. Since the coefficient of t² is negative, the parabola opens downward, so the vertex gives the maximum height. Step 2: Find the time t when maximum height occurs using t = -b/(2a). Here a = -5, b = 15, so t = -15/(2×-5) = -15/-10 = 1.5 seconds. Step 3: Substitute t = 1.5 into the height function: h(1.5) = -5(1.5)² + 15(1.5) + 2 Step 4: Calculate: -5(2.25) + 22.5 + 2 = -11.25 + 22.5 + 2 = 13.25 Step 5: The maximum height is 13.25 meters.

8. Aroha throws a ball whose height is modeled by h(t) = -5t² + 25t + 10. Find the maximum height. Answer: 41.25 Solution: Identify the coefficients: a = -5, b = 25, c = 10 Find the time when maximum height occurs: t = -b/(2a) = -25/(2×-5) = -25/-10 = 2.5 seconds Substitute t = 2.5 into the height function: h(2.5) = -5(2.5)² + 25(2.5) + 10 Calculate (2.5)² = 6.25 Multiply: -5 × 6.25 = -31.25 and 25 × 2.5 = 62.5…
   Full step-by-step solution

   Step 1: Identify the coefficients: a = -5, b = 25, c = 10 Step 2: Find the time when maximum height occurs: t = -b/(2a) = -25/(2×-5) = -25/-10 = 2.5 seconds Step 3: Substitute t = 2.5 into the height function: h(2.5) = -5(2.5)² + 25(2.5) + 10 Step 4: Calculate (2.5)² = 6.25 Step 5: Multiply: -5 × 6.25 = -31.25 and 25 × 2.5 = 62.5 Step 6: Combine all terms: -31.25 + 62.5 + 10 = 41.25 Step 7: The maximum height is 41.25 meters.