At the beginning of a basketball game, the referee tosses the ball vertically into the air. Its height, h, in feet after t seconds is given by h(t) = -16t2 + 24t +5. During what time interval (to the nearest tenth of a second) is the height of the ball greater than 9 feet?
-16t^2 + 24t + 5 > 9
16t^2 - 24t + 4 < 0
3-√5 < 4t < 3+√5
0.2 < t < 1.3
To find the time interval during which the height of the ball is greater than 9 feet, we need to solve the inequality h(t) > 9.
The given equation is h(t) = -16t^2 + 24t + 5.
Substituting h(t) with 9, the inequality becomes -16t^2 + 24t + 5 > 9.
Rearranging the inequality, we get -16t^2 + 24t - 4 > 0.
To solve this quadratic inequality, we can find the roots of the equation -16t^2 + 24t - 4 = 0.
Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / (2a), where a = -16, b = 24, and c = -4.
t = (-24 ± √(24^2 - 4(-16)(-4))) / (2(-16))
t = (-24 ± √(576 - 256)) / (-32)
t = (-24 ± √320) / (-32)
t = (-24 ± 8√5) / (-32)
t = (3 ± √5) / 4
Since the coefficient of t^2 is negative, the graph of the quadratic opens downward. Therefore, the ball will be higher than 9 feet between the roots.
The two roots of the equation are t = (3 + √5) / 4 and t = (3 - √5) / 4.
Thus, the time interval during which the height of the ball is greater than 9 feet is approximately (to the nearest tenth of a second):
[(3 - √5) / 4, (3 + √5) / 4].
To find the time interval during which the height of the ball is greater than 9 feet, we need to solve the inequality h(t) > 9.
Given that the height function is h(t) = -16t^2 + 24t + 5, we can replace h(t) in the inequality:
-16t^2 + 24t + 5 > 9
Now let's rearrange the inequality to solve for t:
-16t^2 + 24t + 5 - 9 > 0
-16t^2 + 24t - 4 > 0
To solve this quadratic inequality, we can use the fact that a quadratic equation is greater than zero when the graph of the equation is above the x-axis. We need to find the values of t for which the graph is above the x-axis, indicating that the height is greater than 9 feet.
To find these values, we can either factorize the quadratic equation or use the quadratic formula. In this case, let's use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In our equation, a = -16, b = 24, and c = -4.
t = (-24 ± sqrt(24^2 - 4 * -16 * -4)) / (2 * -16)
Simplifying further:
t = (-24 ± sqrt(576 - 256)) / -32
t = (-24 ± sqrt(320)) / -32
Now, let's calculate the value inside the square root:
sqrt(320) = 17.888
Substituting this back into the equation:
t = (-24 ± 17.888) / -32
Now we need to find the values of t that make the height greater than 9 feet. Let's solve for t:
t1 = (-24 + 17.888) / -32 = -0.22 (approximately)
t2 = (-24 - 17.888) / -32 = 1.025 (approximately)
Therefore, the time interval during which the height of the ball is greater than 9 feet is approximately -0.2 to 1.0 seconds.