Alright, let's solve this step-by-step.
We are given the height function of the ball:
[tex]\[ h(t) = 2 + 5t - 5t^2 \][/tex]
We need to find the values of [tex]\( t \)[/tex] when the height [tex]\( h \)[/tex] is 3 meters.
So, we set [tex]\( h(t) = 3 \)[/tex]:
[tex]\[ 3 = 2 + 5t - 5t^2 \][/tex]
Let's rearrange this equation to form a standard quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex].
First, subtract 3 from both sides:
[tex]\[ 0 = 2 + 5t - 5t^2 - 3 \][/tex]
Simplify this equation:
[tex]\[ 0 = -5t^2 + 5t - 1 \][/tex]
Now we have a quadratic equation of the form [tex]\( at^2 + bt + c = 0 \)[/tex] where:
[tex]\[ a = -5, \quad b = 5, \quad c = -1 \][/tex]
To solve for [tex]\( t \)[/tex], we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Let's plug in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ b^2 = 5^2 = 25 \][/tex]
[tex]\[ 4ac = 4(-5)(-1) = 20 \][/tex]
[tex]\[ b^2 - 4ac = 25 - 20 = 5 \][/tex]
Now, we will solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-b \pm \sqrt{5}}{2(-5)} \][/tex]
[tex]\[ t = \frac{-5 \pm \sqrt{5}}{-10} \][/tex]
We split this into the two potential solutions:
[tex]\[ t_1 = \frac{-5 + \sqrt{5}}{-10} \][/tex]
[tex]\[ t_2 = \frac{-5 - \sqrt{5}}{-10} \][/tex]
Simplify each solution:
[tex]\[ t_1 = \frac{5 - \sqrt{5}}{10} \][/tex]
[tex]\[ t_2 = \frac{5 + \sqrt{5}}{10} \][/tex]
Now, let's approximate these values and round them to the nearest hundredth:
[tex]\[ t_1 \approx 0.28 \][/tex]
[tex]\[ t_2 \approx 0.72 \][/tex]
Thus, the values of [tex]\( t \)[/tex] for which the ball's height is 3 meters are:
[tex]\[ t = 0.28 \text{ seconds} \quad \text{or} \quad t = 0.72 \text{ seconds} \][/tex]
