To determine when the ball's height is greater than 10 meters, we need to solve the following inequality:
[tex]\[ -4.9 t^2 + 22 t + 0.75 > 10 \][/tex]
First, we simplify the inequality by subtracting 10 from both sides:
[tex]\[ -4.9 t^2 + 22 t + 0.75 - 10 > 0 \][/tex]
[tex]\[ -4.9 t^2 + 22 t - 9.25 > 0 \][/tex]
Next, we solve the quadratic equation
[tex]\[ -4.9 t^2 + 22 t - 9.25 = 0 \][/tex]
to find the critical points. We use the quadratic formula for [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -4.9 \)[/tex], [tex]\( b = 22 \)[/tex], and [tex]\( c = -9.25 \)[/tex]:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substitute the values [tex]\( a = -4.9 \)[/tex], [tex]\( b = 22 \)[/tex], and [tex]\( c = -9.25 \)[/tex]:
[tex]\[ t = \frac{-22 \pm \sqrt{22^2 - 4(-4.9)(-9.25)}}{2(-4.9)} \][/tex]
[tex]\[ t = \frac{-22 \pm \sqrt{484 - 181.3}}{-9.8} \][/tex]
[tex]\[ t = \frac{-22 \pm \sqrt{302.7}}{-9.8} \][/tex]
[tex]\[ t = \frac{-22 \pm 17.4}{-9.8} \][/tex]
Now, calculate the two roots:
[tex]\[ t_1 = \frac{-22 + 17.4}{-9.8} = \frac{-4.6}{-9.8} \approx 0.469 \][/tex]
[tex]\[ t_2 = \frac{-22 - 17.4}{-9.8} = \frac{-39.4}{-9.8} \approx 4.02 \][/tex]
These roots indicate the points where the height is exactly 10 meters. To find when the height is greater than 10 meters, we need to analyze the inequality:
[tex]\[ -4.9 t^2 + 22 t - 9.25 > 0 \][/tex]
Since the quadratic equation [tex]\( -4.9 t^2 + 22 t - 9.25 \)[/tex] opens downwards (as the coefficient of [tex]\( t^2 \)[/tex] is negative), it will be greater than 0 between the two roots. Therefore:
The ball's height is greater than 10 meters when [tex]\( t \)[/tex] is approximately between [tex]\( 0.469 \)[/tex] and [tex]\( 4.02 \)[/tex] seconds.
So, the ball's height is greater than 10 meters when [tex]\( t \)[/tex] is approximately between [tex]\( 0.469 \)[/tex] and [tex]\( 4.02 \)[/tex] seconds.
