Given the model for the height of a ball:
[tex]h(t)=-16t^2+5t+15[/tex]
To hit the ground, the height must be h(t) = 0. Then:
[tex]\begin{gathered} -16t^2+5t+15=0 \\ \Rightarrow16t^2-5t-15=0 \end{gathered}[/tex]
We use the general solution for a quadratic equation:
[tex]ax^2+bx+c=0\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
From the problem, we identify:
[tex]\begin{gathered} a=16 \\ b=-5 \\ c=-15 \end{gathered}[/tex]
Using the formula:
[tex]\begin{gathered} t=\frac{5\pm\sqrt{25+4\cdot15\cdot16}}{2\cdot16}=\frac{5\pm\sqrt{985}}{32} \\ \\ \Rightarrow t_1=1.137 \\ \Rightarrow t_2=-0.825 \end{gathered}[/tex]
We only take the positive value (because the time is always positive!). Then, the answer is:
[tex]1.137\text{ seconds}[/tex]
