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How long is the ball above 7 meters with the equation h = -t^2 + 8t?

A) 4 seconds
B) 6 seconds
C) 8 seconds
D) 10 seconds

User OlivaresF
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1 Answer

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Final answer:

To find out how long the ball is above 7 meters, we solve the quadratic equation -t^2 + 8t = 7, which yields t = 1 second and t = 7 seconds. The ball is above 7 meters for the duration between these two times, which is 6 seconds.

Step-by-step explanation:

The student is asking how long a ball, described by the height equation h = -t^2 + 8t, remains above 7 meters. To determine this, we need to solve the inequality h > 7 for time t using the given equation. To find the timestamps when the height is exactly 7 meters, we solve the equation -t2 + 8t = 7.

Let's rearrange the equation to get the quadratic form: -t2 + 8t - 7 = 0. We can use the quadratic formula, t = (-b ± sqrt(b2 - 4ac))/(2a), where a is the coefficient of t2, b is the coefficient of t, and c is the constant term.

Substituting the values gives us a = -1, b = 8, and c = -7. After calculating, we find the two timestamps when the ball is at exactly 7 meters are t = 1 second and t = 7 seconds. Thus, the ball is above 7 meters between these two times, giving us a duration of 6 seconds.

User Ivayle
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