A ball is thrown vertically upward with an initial velocity of 96 feet per second.The distance s (in feet) of the ball from the ground after t second is s(t)=96t-16t².at what time t is the ball more than 128 feet above the ground?
-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
To find the time at which the ball is more than 128 feet above the ground, we need to solve the equation s(t) > 128.
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.