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The height of a ball is given by the function h(t) = -16t² + 40t + 6, where h(t) is the height of the ball above ground t seconds after the ball is released. How many seconds have passed when the ball reaches the height of 22 feet on its upward path? a) 1.5 seconds b) 2 seconds c) 2.5 seconds d) 3 seconds

User Tcole
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1 Answer

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Answer:

To find the time it takes for the ball to reach a height of 22 feet on its upward path, you can set the height function h(t) equal to 22 and solve for t:

h(t) = 22

-16t² + 40t + 6 = 22

Now, subtract 22 from both sides:

-16t² + 40t + 6 - 22 = 0

Simplify:

-16t² + 40t - 16 = 0

Now, divide the entire equation by -8 to make it easier to work with:

2t² - 5t + 2 = 0

You can solve this quadratic equation using the quadratic formula:

t = (-b ± √(b² - 4ac)) / (2a)

In this equation, a = 2, b = -5, and c = 2.

t = (-(-5) ± √((-5)² - 4 * 2 * 2)) / (2 * 2)

t = (5 ± √(25 - 16)) / 4

t = (5 ± √9) / 4

t = (5 ± 3) / 4

Now, consider both solutions:

1. t = (5 + 3) / 4 = 8 / 4 = 2 seconds

2. t = (5 - 3) / 4 = 2 / 4 = 0.5 seconds

Since you're interested in the time it takes for the ball to reach a height of 22 feet on its upward path, you should choose the positive value:

The ball reaches a height of 22 feet on its upward path after 2 seconds.

So, the correct answer is (b) 2 seconds.

Explanation:

User Pkmmte
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