-16t² + 96t - 128 = 0 ... t² - 6t + 8 = 0
solve for t ... the two times are upward and downward
...the ball is above 128 ft between the times
solve for t ... the two times are upward and downward
...the ball is above 128 ft between the times
Given that s(t) = 96t - 16t², we can set up the inequality 96t - 16t² > 128.
To solve this inequality, we can first rearrange it to get a quadratic in standard form, setting it equal to zero:
-16t² + 96t - 128 > 0
Now, let's solve this quadratic inequality using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
In this case, a = -16, b = 96, and c = -128. Plugging these values into the formula, we get:
t = (-96 ± √(96² - 4(-16)(-128))) / (2(-16))
Simplifying:
t = (-96 ± √(9216 - 8192)) / (-32)
t = (-96 ± √1024) / (-32)
Now we have two possible solutions:
1. t = (-96 + √1024) / (-32)
2. t = (-96 - √1024) / (-32)
Simplifying further:
1. t = (-96 + 32) / (-32)
t = -64 / (-32)
t = 2
2. t = (-96 - 32) / (-32)
t = -128 / (-32)
t = 4
Therefore, the ball is more than 128 feet above the ground at t = 2 seconds and t = 4 seconds.