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A football is kicked into the air. Its height above the ground changes according to the equation y --16x2 + 80x +3, where x = time in seconds and ) = height in feet. The ball reaches a height of

67 feet twice: once on the way up and once on the way down. How much time does the ball spend at or above a height of 67 feet?

User Sharesse
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1 Answer

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Final answer:

The football spends a total of 3.25 seconds at or above 67 feet. This is found by solving the quadratic equation for the time when the football is at 67 feet, resulting in two-time values, 3.79 seconds and 0.54 seconds. The difference between these times gives the duration above 67 feet.

Step-by-step explanation:

To calculate the time the football spends at or above a height of 67 feet, we need to find when the height (y) equals 67 feet using the given quadratic equation: y = -16x2 + 80x + 3. We set this equation equal to 67 and solve for x using the quadratic formula. Here's the adjusted equation:

67 = -16x2 + 80x + 3.

First, we rearrange the equation: 0 = -16x2 + 80x - 64. Now we apply the quadratic formula x = (-b ± √(b2 - 4ac)) / (2a), where a = -16, b = 80, and c = -64. Using these values, we calculate the two possible solutions for x.

After solving, we find two-time values that correspond to when the ball is at 67 feet: t = 3.79 s and t = 0.54 s. To find the total time the ball spends at or above 67 feet, we simply subtract the smaller time from the larger time:

Total time at or above 67 feet = 3.79 s - 0.54 s = 3.25 seconds.

User Hezye
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