Respuesta :
The equation representing the ball's movement is given as follows;
[tex]h=217-17t-16t^2[/tex]To determine the value of t (that is, time), we shall now make the equation become;
[tex]217-17t-16t^2=0[/tex]This is because when the height is zero, then the ball has actually touched the ground. So making h equal to zero, we now have a quadratic equation and we shall solve it using the quadratic equation formula as follows;
[tex]t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]The variables are,
[tex]\begin{gathered} -16t^2-17t+217=0 \\ a=-16,b=-17,c=217 \end{gathered}[/tex]We now have;
[tex]\begin{gathered} t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t=\frac{-(-17)\pm\sqrt[]{(-17)^2-4\lbrack-16\rbrack\lbrack217\rbrack}}{2(-16)} \\ t=\frac{17\pm\sqrt[]{289+13888}}{-32} \\ t=\frac{17\pm\sqrt[]{14177}}{-32} \\ t=\frac{17\pm119.0672}{-32} \\ t=\frac{17+119.0672}{-32},t=\frac{17-119.0672}{-32} \\ t=-4.2521,t=3.1896 \end{gathered}[/tex]The quadratic equation now has two solutions.
Note however, that the question requires us to determine the time taken and we know this cannot be a negative value.
Therefore, we shall take t = 3.1896, and rounded to the nearest hundredth this becomes;
[tex]t\approx3.19\text{ (rounded to the nearest hundredth)}[/tex]