To determine when the ball is more than 128 feet above the ground, we will use the formula for the height \( h(t) \) of the ball at time \( t \):
\[ h(t) = -16t^2 + vt + h_0 \]
In this case, the initial velocity \( v \) is 96 feet per second and the initial height \( h_0 \) is 0 (since the ball is kicked from the ground). Thus, the equation becomes:
\[ h(t) = -16t^2 + 96t \]
We need to find when \( h(t) > 128 \):
\[ -16t^2 + 96t > 128 \]
Rearranging this, we subtract 128 from both sides:
\[ -16t^2 + 96t - 128 > 0 \]
Next, we can simplify the inequality by dividing everything by -16 (remember to reverse the inequality sign):
\[ t^2 - 6t + 8 < 0 \]
Now, we need to factor the quadratic:
\[ (t - 2)(t - 4) < 0 \]
To solve this inequality, we find the roots of the equation \( (t - 2)(t - 4) = 0 \), which gives us:
\[ t = 2 \quad \text{and} \quad t = 4 \]
Now we analyze the intervals determined by these roots:
- \( (-\infty, 2) \)
- \( (2, 4) \)
- \( (4, \infty) \)
We test each interval:
- For \( t < 2 \) (e.g., \( t = 1 \)): \((1 - 2)(1 - 4) = (-1)(-3) = 3 > 0\)
- For \( 2 < t < 4 \) (e.g., \( t = 3 \)): \((3 - 2)(3 - 4) = (1)(-1) = -1 < 0\)
- For \( t > 4 \) (e.g., \( t = 5 \)): \((5 - 2)(5 - 4) = (3)(1) = 3 > 0\)
The inequality \( (t - 2)(t - 4) < 0 \) holds true in the interval \( (2, 4) \).
Thus, the ball is more than 128 feet above the ground between 2 seconds and 4 seconds.
Final answer: The ball is more than 128 feet above the ground between 2 and 4 seconds.
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