Answer:
To find the values of \( t \) for which the ball's height \( h \) is 8 meters, we need to set the equation for \( h \) equal to 8 and solve for \( t \):
\[ h = 3 + 11t - 5t^2 \]
Setting \( h \) to 8 meters:
\[ 8 = 3 + 11t - 5t^2 \]
Now we need to solve the quadratic equation:
\[ 5t^2 - 11t + (3 - 8) = 0 \]
\[ 5t^2 - 11t - 5 = 0 \]
This is a quadratic equation in the form \( at^2 + bt + c = 0 \), where \( a = 5 \), \( b = -11 \), and \( c = -5 \). We can solve this equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \) into the formula:
\[ t = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(5)(-5)}}{2(5)} \]
\[ t = \frac{11 \pm \sqrt{121 + 100}}{10} \]
\[ t = \frac{11 \pm \sqrt{221}}{10} \]
Now, calculate the square root of 221 and then the two possible values for \( t \):
\[ \sqrt{221} \approx 14.866 \]
The two possible solutions for \( t \) are:
\[ t = \frac{11 + 14.866}{10} \]
\[ t = \frac{25.866}{10} \]
\[ t \approx 2.59 \] (rounded to the nearest hundredth)
And
\[ t = \frac{11 - 14.866}{10} \]
\[ t = \frac{-3.866}{10} \]
\[ t \approx -0.39 \] (rounded to the nearest hundredth)
Since time cannot be negative in this context, the negative value of \( t \) does not make sense for the problem. Therefore, the only physically meaningful solution is \( t \approx 2.59 \) seconds.
Explanation: