To determine the values of \( t \) for which the ball's height \( h \) is non-negative (i.e., \( h \geq 0 \)), we need to use the given height equation:
\[ h = 2 + 9t - 5t^2 \]
We want to solve the inequality:
\[ 2 + 9t - 5t^2 \geq 0 \]
This is a quadratic inequality. First, let's rewrite it in standard form:
\[ -5t^2 + 9t + 2 \geq 0 \]
Now, for easier calculations, we can multiply by \(-1\) (remembering to reverse the inequality sign):
\[ 5t^2 - 9t - 2 \leq 0 \]
Next, we will find the roots of the quadratic equation \( 5t^2 - 9t - 2 = 0 \) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 5 \), \( b = -9 \), and \( c = -2 \):
\[ t = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \cdot 5 \cdot (-2)}}{2 \cdot 5} \] \[ t = \frac{9 \pm \sqrt{81 + 40}}{10} \] \[ t = \frac{9 \pm \sqrt{121}}{10} \] \[ t = \frac{9 \pm 11}{10} \]
Calculating both roots:
- \( t = \frac{20}{10} = 2 \)
- \( t = \frac{-2}{10} = -0.2 \)
The roots are \( t = 2 \) and \( t = -0.2 \).
Next, we need to determine the intervals for which the quadratic is less than or equal to 0. The critical points divide the real line into intervals:
- \( (-\infty, -0.2) \)
- \( (-0.2, 2) \)
- \( (2, \infty) \)
To test the sign of \( 5t^2 - 9t - 2 \) within each interval, we can select a test point from each interval:
-
For the interval \( (-\infty, -0.2) \): Choose \( t = -1 \): \[ 5(-1)^2 - 9(-1) - 2 = 5 + 9 - 2 = 12 > 0 \]
-
For the interval \( (-0.2, 2) \): Choose \( t = 0 \): \[ 5(0)^2 - 9(0) - 2 = -2 < 0 \]
-
For the interval \( (2, \infty) \): Choose \( t = 3 \): \[ 5(3)^2 - 9(3) - 2 = 45 - 27 - 2 = 16 > 0 \]
Thus, the quadratic \( 5t^2 - 9t - 2 \) is less than or equal to zero in the interval \( (-0.2, 2) \).
Finally, we also include the endpoints where the quadratic is equal to zero:
Thus, the values of \( t \) for which the ball's height is non-negative are:
\[ t \in [-0.2, 2] \]