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Math  /  Geometry

QuestionA Longhorn kicker recently kicked a 54-yard field goal during practice. The football started at a height of 0 ft , and the football was 12 ft off the ground when it made it a horizontal distance of 54 yards. Assume that the football hits the ground 10 yards after it passes the goal post. What was the maximum height of the football? Round to the nearest whole number in FEET (1 yard = 3 feet). □\square

Studdy Solution

STEP 1

1. The football's trajectory can be modeled by a quadratic function.
2. The football starts at a height of 0 feet.
3. The football is 12 feet off the ground at a horizontal distance of 54 yards.
4. The football hits the ground 10 yards after it passes the goal post.
5. 1 yard is equivalent to 3 feet.

STEP 2

1. Convert all distances from yards to feet.
2. Set up a quadratic equation to model the football's trajectory.
3. Use the vertex form of a quadratic equation to find the maximum height.

STEP 3

Convert all given distances from yards to feet:
- The football is 54 yards from the start when it is 12 feet high. 54 yards×3 feet/yard=162 feet 54 \text{ yards} \times 3 \text{ feet/yard} = 162 \text{ feet}
- The football hits the ground 10 yards after passing the goal post. (54+10) yards×3 feet/yard=192 feet (54 + 10) \text{ yards} \times 3 \text{ feet/yard} = 192 \text{ feet}

STEP 4

Set up a quadratic equation to model the football's trajectory:
The general form of a quadratic equation is: y=ax2+bx+c y = ax^2 + bx + c
Given points:
1. (0,0) (0, 0) - the starting point.
2. (162,12) (162, 12) - the point where the football is 12 feet high.
3. (192,0) (192, 0) - the point where the football hits the ground.
Use these points to find the coefficients a a , b b , and c c .
Since c=0 c = 0 (the starting height is 0), the equation simplifies to: y=ax2+bx y = ax^2 + bx
STEP_2.1: Substitute the points into the simplified equation to create a system of equations:
1. For (162,12) (162, 12) : 12=a(162)2+b(162) 12 = a(162)^2 + b(162)
2. For (192,0) (192, 0) : 0=a(192)2+b(192) 0 = a(192)^2 + b(192)
Solve this system of equations for a a and b b .
STEP_2.2: Solve the system of equations:
From the second equation: 0=a(192)2+b(192) 0 = a(192)^2 + b(192) 0=36864a+192b 0 = 36864a + 192b b=−192a b = -192a
Substitute b=−192a b = -192a into the first equation: 12=a(162)2+(−192a)(162) 12 = a(162)^2 + (-192a)(162) 12=26244a−31104a 12 = 26244a - 31104a 12=−4860a 12 = -4860a a=−124860 a = -\frac{12}{4860} a=−1405 a = -\frac{1}{405}
Substitute a a back to find b b : b=−192(−1405) b = -192(-\frac{1}{405}) b=192405 b = \frac{192}{405}

STEP 5

Use the vertex form of the quadratic equation to find the maximum height:
The vertex form is: y=a(x−h)2+k y = a(x - h)^2 + k
The vertex (h,k) (h, k) is the maximum point for a downward-opening parabola. The vertex h h can be found using: h=−b2a h = -\frac{b}{2a}
Substitute a=−1405 a = -\frac{1}{405} and b=192405 b = \frac{192}{405} : h=−1924052(−1405) h = -\frac{\frac{192}{405}}{2(-\frac{1}{405})} h=96 h = 96
Substitute x=96 x = 96 back into the quadratic equation to find k k : y=−1405(96)2+192405(96) y = -\frac{1}{405}(96)^2 + \frac{192}{405}(96) y=−9216405+18432405 y = -\frac{9216}{405} + \frac{18432}{405} y=9216405 y = \frac{9216}{405}
Convert to a whole number: y≈23 y \approx 23
The maximum height of the football is:
23 feet \boxed{23 \text{ feet}}

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